GT Cau truc du lieu

8/3/2019 GT Cau truc du lieu

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Cu trc d liu v gii thut Ng Hu Phc

MN HCCU TRC D LIU V GII THUT

Bi 1: GII THIU V THUT TON V CU TRC D LIU1.1. Khi nim thut ton

1.1.1. nh ngha:

Thut ton l mt s hu hn cc bc thc hin tnh ton theo mt trnh t

xc nh bin i cc gi tr u vo (Input) thnh cc gi tr u ra mong mun

(Output).

VD: Bi ton gii phng trnh bc nht ax + b=0

Thut ton:Kim tra gi tr ca a;

Nu =0 th

Nu b=0 th VSN

Ngc li th VN

Ngc li

X=-b/a;

1.1.2. Cc c trng ca thut ton

1. u vo (Input), ra (Output) : mi thut ton cn phi c b d liu u

vo v d liu u ra (kt qu tnh ton).

2. Tnh xc nh: mi bc ca thut ton cn phi m t chnh xc, r

rng.

3. Tnh kh thi: tt c cc php ton trong thut ton phi n gin

thc thi c.

4. Tnh dng: vi mi b s liu u vo tha mn iu kin ban u th

thut ton phi dng sau hu hn bc.

5. Tnh ph dng: Thut ton c th p dng i vi cc d liu u vo

khc nhau trong mt min xc nh.

1.2. Biu din thut ton.

1. Biu bng cch lit k cc bc

Lit k ra cc bc cn phi thc hin bng ngn ng t nhin.VD 1: Thut ton tm s ln nht trong 3 s a,b,c

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Gi thit s ln nht l a: Max=a

Nu Max


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Cu trc d liu v gii thut Ng Hu Phc Nt ch cc iu kin c dng hnh thoi. Trong cc ng ni vi

nt iu kin ny c hai ng i ra ng vi hai trng hp iu kin nghoc iu kin sai:

Khi hnh bnh hnh in cc gi tr. Cc ng ni cc nt vi nhau.

V d 1: s khi sau y m t thut ton tm gi tr ln nht trong mng s A

c n phn t. Thut ton ny c m t trong v d 1.

S khi thut ton tm max trong mng A

3. Biu din thut ton bng m gi

Vi nhng thut ton phc tp, vic v v theo di s khi thng l khngthun tin. Trong mt s trng hp ngi ta s dng ngn ng gi m (gi ngnng lp trnh) m t thut ton.

Cc cu trc c bn c dng trong ngn ng gi m cng nh trong bt kngn ng no khc l cu trc tun t, cu trc r nhnh v cu trc lp.

a) Cu trc tun t: Lit k cc cng vic, cc thao tc theo th t. h tr cho vic theo di c thun tin cng c th thm s th t.

b) Cu trc r nhnh:

If (iu kin) Then (cng vic);

If (iu kin) Then (cng vic 1) Else (cng vic 2);

c) Cu trc lp:

For (bin):= (gi tr u) to (gi tr cui) do (cng vic); For (bin):= (gi tr u) downto (gi tr cui) do (cng

vic);

While (iu kin) do (cng vic);

Repeat- (cng vic 1);- (cng vic 2);- ....

- (cng vic k);Until (iu kin);

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VD: Thut ton tm s ln nht trong 3 s a,b,c

*) Lit k cc bc:

Max:=a;

If Max


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Cu trc d liu v gii thut Ng Hu Phcelse

return F(n-1) + F(n-2);}

Cch thc hat ng nh sau:

F4/ \

F3 F2/ \ / \

F2 F1 F1 F0/ \

F1 F0Trong v d trn, ta thy ta phi tnh 2 ln F2, 3 ln F1 v 2 ln F0.

Do phng php ny chy rt chm khi n ln, thm ch c th b trnstack.

Cc thut ton Quick Sort, Merge Sort thit k theo phng php ny

1.3.2. QUY HOCH NG.

Nguyn l quy hoch ng cng da theo nguyn l chia tr nhng khc

ch l phng php ny tip cn theo hng bottom-up (t di ln).u tin, chng ta xut pht t nhng bi ton c s, d dng c ngay li gii.

T tp nhng li gii ca cc bi ton con nh th, ta tm c li gii ca bi ton

ln hn. V c thc hin nh th cho n khi thu c bi ton cui cng mong

mun.

VD: Xt l bi ton s FibonaciBi ton trn c gii theo phng php quy hoch ng nh sau:

int F(int n){int Fi[100];int i;Fi[0] = Fi[1] = 1;for (i = 2; i


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Cu trc d liu v gii thut Ng Hu Phc{

int f, f1, f2, i;f1 = f2 = 1;for (i = 2; i


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Cu trc d liu v gii thut Ng Hu Phcthi im ch c th thc hin mt cng vic. Mi cng vic i c thi gian bt u lsi v thi gian kt thc l i, vi si i.

Nu c chn, cng vic i chim ht khong thi gian [si,i].

Hy chn ra phng n thc hin cc cng vic trn sao cho s cng vic thc

hin c l ln nht trong mt khong thi gian nht nh.Thut ton tham lam: u tin ta sp xp cc cng vic trn tng dn theo thi

gian kt thc.

Thc hin cng vic u tin.

I=1

Cho j=i+1 to n

Nu sj>ai th

Chn tip j

I=j;

p dng k thut tham n s cho mt gii thut thi gian a thc, tuy nhin ni

chung chng ta ch t c mt phng n tt ch khng phi l ti u.

1.3.5. K THUT QUI v QUI QUAY LUI

VD : Bi ton xp 8 hu ln bn c.

1.4. Quan h gia cu trc d liu v thut ton.

- Thut ton ch ch ra cc bc tnh ton, cn i tng c dng tnh

ton chnh l d liu.

- D liu biu din cc thng tin cn thit cho bi ton: cc d liu u vo,

trung gian, kt qu. Nu bit t chc d liu theo cu trc thch hp th vic x l s

thun tin hn v t c hiu qu cao hn.

- Vi mt cu trc d liu chn ta s c gii thut tng ng. Cu trc d

liu thay i th gii thut cng thay i theo.

Xt v d sau: Gi s u vo chng ta c mt cp danh sch: Tn n v, sin thoi. (a1,b1),(a2,b2), Ta mun c chng trnh tm s in thoi theo tn

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n v.

Gii thut 1: Tm kim tun t t u danh sch tn cc n v c trong danh

sch cho n khi tm thy hoc ht danh sch.

Gii thut 2: Trc khi tm kim, ta t chc li d liu bng cc sp xp danh

sch tng dn theo t in. Vic tm kim s thc hin t u danh sach cho n khi

tm thy hoc tnnvcntm


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VD3: Ch ra hm n! l O(nn) v log(n!) l O(nlogn)

Ta c n!


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Bi 2 : QUY

2.1. Khi nim v quy.

Ta ni mt i tng l qui nu i tng ny bao gm chnh n nh mt

b phn hoc i tng c nh ngha di dng ca chnh n.

V d 1: giai tha

+ 0!=1

+ n !=n*(n-1) !

V d 2 : Cu trc th mc l mt dng nh ngha qui : th mc cha th

mc con.

2.2. Gii thut qui v th tc qui- Nu li gii ca bi ton T c thc hin thng qua li gii ca mt bi ton

T c kch thc nh hn T v c dng ging nh T, th l mt li gii qui.

Gii thut tng ng vi li gii nh vy gi l gii thut qui.

- Trong th tc c li gi n chnh n gi l th tc qui.

2.3. Thit k gii thut quy

Khi bi ton c nh ngha di dng qui th ta rt d thit k gii thut

.V d 1 : Hm giai tha n !

nh ngha n!

n! =1 nu n=0

= n(n-1) ! nu n>0

#include

int f(int n)

{ if (n==0) return 1;

else

return n*f(n-1);

}

void main()

{ int fb;

fb=f(4);

printf("F=%d",fb);

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}

v d 2 : Dy Fibonacci

fib(n)=fib(n-2)+fib(n-1) nu n>2

fib(n)=1; nu n


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4. Tng t ta dng mng dcp vi ngha: dcp[k]=1 nu ng cho ph th

k c mt con hu c t.

Gi m ca thut ton xp hu nh sau:

procedure Try(i);

var j;

begin

for j := 1 to n do

if (cot[j]=0) and (dcc[i-j]=0) and (dcp[i+j]=0) then

begin

x[i] := j; cot[j]:=1; dcc[i-j]:=1; dcp[i+j]:=1;{ghi nhn trng thi mi}

if i=n then Updateelse Try(i+1);

cot[j]:=0; dcc[i-j]:=0; dcp[i+j]:=0; {phc hi trng thi c}

end;

end;

procedure Update;

begin

count := count + 1;print(x);

end;

{tng cng c 92 cch xp 8 hu}

2.4. Kh qui.

C nhng bi ton bn cnh gii thut qui vn c nhng gii thut lp kh

n gin v hiu qu.

Vd 1: n!

int F(int n)

{int tg;

if ((n==0)||(n==1)) { return 1; }

else

{ tg=1;

for (int i=2;i


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return tg;

} }

VD2: Fibonacci

Function Fibo (n:byte):double;

Begin

If n


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2.6. Tnh ng n ca cc gii thut quy

chng minh tnh ng n ca mt gii thut qui ngi ta s dng

phng php qui np ton hc.

Vd: Chng minh gii thut qui tnh giai tha F(n).

Gii thut:

F(0)=1

F(n)=n(n-1)(n-2)2*1;(n>0)

CM:

+Nu n=0; theo gii thut f:=1 (nh vy gii thut ng khi n=0)

+ Gi s gii thut ng vi n=k, ngha l F(k)=k*(k-1)*2*1.

+ Ta phi chng minh gii thut ng vi n=k+1;Ta thy vi n=k+1 th F:=n*f(n-1) s c thc hin. Thay n=k+1 ta c

F(k+1)=(k+1)F(k)=(k+1)*k*(k-1)*2*1.(dpcm)

2.7. Bi tp

1) Xt tnh qui:

n+1 nu m=0

A(m,n) = A(m-1,1) nu n=0A(m-1,A(m,n-1)) vi cc trng hp cn li.

2) Bi ton 8 qun hu v gii thut quay lui

Bn c vua gm 8 hng, 8 ct. Qun hu l qun c c th n c bt k qun

no nm trn cng hng, ct, ng cho. Bi ton: hy xp 8 con hu ln trn bn

c sao cho cc qun hu khng th n c nhau.

3) Din tch bn c

Cho bng ch nht A gm M dng v N ct. Trn mi ghi cc gi tr 0 v 1:

gi tr 0 c hiu l nc, gi tr 1 c hiu l t. Hai t c cnh lin nhau

c xem nh nm trong cng mt bn c. Hy xc nh din tch ca bn c ln

nht (mi c xem nh 1 n v o).

Nhn xt:

- Mi c nh s 1 c xem nh l mt nh ca th.

- Hai nh c gi l lin thng nu c chung cnh. Nh vy yu cu ca bi

ton chnh l tm thnh phn lin thng c s nh ln nht.

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Bi 3: MNG V DANH SCH.

3.1. Cc khi nim.

Mng: l mt tp hp c th t cc phn t. Vector c lu trong mng mt

chiu, mi phn t ca n ng vi mt ch s i. Ma trn c lu trong mt mng 2

chiu, mi phn t ng vi 2 ch s i,j. S phn t ca mng l c nh, c xc

inh khi khai bo.

Danh sch: l mt tp hp c th t cc phn t. S phn t ca danh sch

khng c nh, c th thm hoc xa i.

3.2. Cc thao tc trn mng.

3.2.1. Truy cp cc phn t ca mng.Trong C++, khai bo mng nh sau:

type name [elements]; VD: int billy [5];Khi to mng: int billy [5] = { 16, 2, 77, 40, 12071 };Cc phn t ca mng trong C++ c nh s t 0:

Truy cp n phn t ca mng: Billy[2]=100; x=Billy[1];Mng nhiu chiu: int jimmy [3][5];

jimmy[n][m]=(n+1)*(m+1);

3.2.2. Chn thm phn t vo gia mng.

chn mt phn t vo mng, ta cn phi xt xem mng cn ch trng

khng? Nu cn, ta s y cc phn t t v tr cn chn i mt nh v cui mng

ri t gi tr mi vo v tr cn chn.V d:

a[1] a[2] a[3] a[4] a[5] a[6]

Gi s ta mun chn vo v tr s 4 ca mng ta thc hin nh sau:a[1] A[2] a[3] a[4] a[5] a[6] a[7]

Thut ton:

Cc tham s: mng A; S phn t, Kch thc mng, v tr chn, gi tr cn

chn. void Insert(int *A, int N , int Size, int k, int x){

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Cu trc d liu v gii thut Ng Hu Phcif (N


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3.3. DANH SCH LIN KT.

3.3.1. Mt s php ton trn danh sch

1. Khi to danh sch rng:Initialize(var L: List)

2. Xc nh di ca danh sch:Length(L: List)

3. Loi phn t v tr th p ca danh sch: Delete(var L:List, p:

position)

4. Chn phn t x vo v tr p ca danh sch: Insert(var L: List, p:

position, x: item)

5. Tm x c trong danh sch L khng? Search(x:item, L: List, var p:

position)

6. Kim tra danh sch c rng khng?IsEmpty(L:List)

3.3.2. Khi nim v cch s dng con tr.

Khi nim

Con tr l mt khi nim c trong mt s ngn ng lp trnh nh c, c++,

pascal dng qun l b nh ng.

Bin tnh: l bin c xc nh vng nh khi chng trnh bt u chy v

b hy khi chng trnh kt thc.

Bin ng: l bin c to ra v c th hy b khi chng trnh ang thc

hin.

Bin con tr dng qun l b nh ng.

Khai bo v s dng con tr C++

- Khai bo:

type * pointer_name;

VD:

int * i;char * c; float * f;

- Cp pht vng nh v dng con tr p qun l

Pointer_name = new type ([Init_value])

int *i;

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i= new int(1012);

printf("i=%d",*i);

- Thao tc trn vng nh cp pht thng qua con tr p

+ Gn gi tr cho vng nh: *pointer_name:=x;

+ Ly gi tr ca vng nh: y:=*pointer_name;

+ Hai con tr cng kiu c th thc hin php gn, php so snh

q=p;==, !=,>=,...

- Hy vng nh

deletepointer_name;

3.3.3. Danh sch lin kt n.

Mi phn t ca danh sch n l mt cu trc cha 2 thng tin:

- Thnh phn d liu: lu tr cc thng tin v bn thn phn t .

- Thnh phn mi lin kt: lu tr a ch ca phn t k tip trong danh

sch, hoc lu tr gi tr NULL nu l phn t cui danh sch.

Cch qun l: Nu bit c a ch ca phn t u tin trong danh sch n

th c th da vo con tr Next truy xut n phn t th 2 trong xu, v li da

vo con tr Next ca phn t th 2 truy xut n phn t th 3...

Khai bo tng qut

struct NODE{

int Info;

struct NODE * pNext;};

NODE* Head;NODE *xNew;int x;

18

aan

Head


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Khi to danh sch rng: Head=NULL;

To mt phn t cho danh sch

NODE* GetNode(int x)

{ NODE *p;

p=new NODE();

if (p==NULL)

{

printf("Khng ?? b? nh?.");

return NULL;

}

p->Info=x;

p->pNext=NULL;

return p;

}

Kim tra danh sch c rng hay khng

bool Empty(NODE *h)

{

return (h==NULL);

}

Thm mt phn t vo u danh sch

Thut ton thm phn t p vo u danh sch :

Bt u:

B11 : p=GetNode(x); p->Next = Head;B12 : Head = p ;

an-1

a1

an

X

Head

an-2

p

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NODE* InsertHead(NODE **h, int x)

{ NODE* xNew = GetNode(x);

if (xNew ==NULL) return NULL;

if (*h==NULL)

{

*h = xNew;

}

else

{

xNew->pNext = *h;

*h = xNew;}

return xNew;

}

Chn mt phn t vo sau phn t q trong danh sch

Thut ton chn phn t p:

Bt u :

Nu ( p != NULL) th

B1 : p -> Next = q->Next;

B2 : q->Next = p ;

NODE * InsertAfter(NODE **q, int x)

{

NODE* new_ele = GetNode(x);

x

a1an-1an

Head

an-2

q

p

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if (new_ele ==NULL) return NULL;

if ( *q!=NULL)

{

new_ele->pNext = (*q)->pNext;

(*q)->pNext = new_ele;

}

else //chn vo ??u danh sch

InsertHead(&Head, x);

}

Chn phn t vo cui danh sch

Thut ton chn p vo cui danh sch:

Bt u :Nu danh sch rng th

Head = p;

Ngc li

B21 : q:=head;

B22 : trong khi p nil th

Before:=q;q:=q^.next;

B23:before^.next:=p;

NODE* InsertTail(NODE **h, int x)

{

NODE* new_ele = GetNode(x);

if (new_ele ==NULL) return NULL;

if (*h==NULL)

{

an-1

a1

an

X

Head

an-2

pq

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*h = new_ele;

}

else

{

NODE *p;

p=*h;

while (p->pNext !=NULL) p=p->pNext ;

p->pNext = new_ele;

}

return new_ele;

}

Th tc xa mt phn t ra khi danh sch

Hy phn t u xu:

Bt u:

Nu (Head != NULL) th

B1: p = Head; // p l phn t cn hy

B2:

B21 : Head = Head->Next; // tch p ra khi xu

B22 : delete p; // Hy bin ng do p tr n

Pr

bool RemoveHead(NODE **h){

NODE *p;

if ( *h != NULL){

p = *h;*h= (*h)->pNext;delete p;

an-2

a1

an-1

an

Head

an-3

P q

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}return true;

}

Hy mt phn t ng sau phn t q

Thut ton :

Bt u:

Nu (q!= NULL) th

B1: p = q->Next; // p l phn t cn hy

B2: Nu (p != NULL) th // q khng phi l cui xu

B21 : q->Next = p->Next; // tch p ra khi xu

B22 : dispose(p); // Hy bin ng do p tr n

bool RemoveAfter ( NODE *q)

{ NODE *p;

if ( q != NULL)

{ p = q ->pNext ;

if ( p != NULL)

{ q->pNext = p->pNext;delete p;

}

}

}

Hy mt phn t c kho k Thut ton :

Bc 1:Tm phn t p c kha k v phn t q ng trc n

Bc 2:

Nu (p!= NULL) th // tm thy k

Hy p ra khi danh sch tng t hy phn t sau q;

Ngc li

Bo khng c k;

bool RemoveNode(NODE *h, int k){

NODE *p = h;

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Cu trc d liu v gii thut Ng Hu PhcNODE *q = NULL;

while( p != NULL){

if(p->Info == k) break;q = p; p = p->pNext;

}

if(p == NULL) return false; //Khng tm th?y k

if(q != NULL){

q->pNext = p->pNext;delete p;

}else //p l ph?n t? ??u xu{

RemoveHead(&Head);}

return true;}

X l danh sch

- m cc phn t ca danh sch,

- Tm phn t tho iu kin,

- Hu ton b danh sch (v gii phng b nh) duyt danh sch (v x l tng phn t) ta thc hin cc thao tc sau:

Thut ton :

Bc 1:

p = Head; //Cho p tr n phn t u danh sch

Bc 2:

Trong khi (Danh sch cha ht) thc hin

B21 : X l phn t p;

B22 : p:=p->Next; // Cho p tr ti phn t k

Hy ton b danh sch void RemoveList(NODE **h){ NODE *p;

while ( (*h) != NULL){

p=*h;(*h) = (*h)->pNext;

delete p;

}

}

In tan b danh schvoid ProcessList (NODE *h)

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Cu trc d liu v gii thut Ng Hu Phc{ NODE *p;

p = h;while (p!= NULL){

printf("%d == ",p->Info );p = p->pNext;

}printf("\n");

}Tm kim

NODE *Search(NODE *h, int k){

NODE *p;p = h;

while((p!= NULL)&&(p->Info != k))p = p->pNext;

return p;

}

3.3.4. Cc loi danh sch lin kt khc.

Danh sch lin kt vng

Con tr ca phn t u tin tr v phn t t cui danh sch.

Khi x l danh sch ni vng nu khng cn thn s dn ti mt chu trnh (ko

kt thc c). d dng cho vic x l ngi ta thng a vo mt nt c bit

gi l nt u danh sch khng cha d liu v con tr pNext tr n chnh n.

struct NODE{ int info;

Struct NODE *pNext;

}

Khi to:

NODE *Head;

Head=new NODE(); Head->info=0; Head->pNext=Head;

InsertHead

p=GetNode(x);p->pNext=Head->pNext;

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an-1an

Head


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Head->pNext=p;

DeleteHead

p=Head->pNext;

if (p!=NULL)

{

Head->pNext=p->pNext;

Delete p;

}

Danh sch lin kt i

Danh sch lin kt i ni vng

3.4. Bi tp.

1. Thit k cu trc d liu biu din ma trn tha.

2. S dng danh sch lin kt vng gii bi ton Joesphus.3. S dng cu trc danh sch biu din a thc hai bin s x v y.

4. Vit hm m s phn t trong mt danh sch lin kt ni vng.

5. Vit mt hm o ngc lin kt ca mt danh sch lin kt n.

6. Sp xp danh sch lin kt n theo trt t tng dn.

7. S dng cu trc danh sch lin kt biu din hai a thc P(x) v Q(x).

a. Vit th tc cng hai a thc ny

b. Vit th tc nhn hai a thc ny

LinkH s

S m

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Bi 4. NGN XP V HNG I.

4.1. Ngn xp.

4.1.1. Khi nim v ngn xp.

Ngn xp (Stack) l mt dng ca danh sch lin kt hot ng theo nguyn

tc vo trc ra sau (First in last out- FILO). Phn t ph trn gi l nh ngn xp,

pha di gi l y ngn xp.

- V d: p dng Stack trong bi ton i c s ca mt s t h thp phn

sang h nh phn.

(m t thut ton i c s ng dng cu trc d liu Stack cho thut ton ny )V d : (10)10=(1010)2

4.1.2. Ci t Stack bng mng.

Ta c th to mt stack bng cch khai bo mt mng 1 chiu vi kch thc

ti a l N (v d, N c th bng 1000).VD:

To stack S v qun l nh stack bng bin t ch s ca phn t trn cng

trong stack:

Khao bo int S[N]; t=-1;B sung vo Stack void Push (int S[], t,x)

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0 0

1

01


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if (t=0)

(Pop=S[t];t-=1;)

4.1.3. Ci t ngn xp bng danh sch lin kt.

Khai bo

struct NODE

{ int info;

struct NODE *pNext;

};

NODE *S;

1. Khi to

S=NULL;

2. Kim tra Stack rng

bool IsEmpty(NODE *s)

{

return (s==NULL);

}

3. Thm phn t mi

void Push(NODE **s, int x)

{NODE *tg;tg=GetNode(x);

tg->pNext=*s;

*s=tg;

}

4. Ly ra mt phn t

int Pop(NODE **s)

{NODE *tg;

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int x;

if (!IsEmpty(*s))

{

x=(*s)->info;

tg=*s;

*s=(*s)->pNext;

delete tg;

return x;

}

return 0;

}4.1.4. ng dng ca ngn xp: tnh gi tr ca biu thc

1. Biu thc trung t: a*(b+c)-d/e

2. Bin i biu thc trung t sang dng hu t

Biu thc s dng k php Balan l mt biu thc khng c ngoc php ton

trong biu thc Balan t sau cc ton hng:

Dng trung t

a+b

a*b

a*(b+c)-d/e

Dng hu t

ab+

ab*

abc+*de-/

Di y l thut ton bin i t dng trung t sang dng hu t

V d :

E=a*(b+c)-d/e#E S E1

$a $ a* $* a( $*( a

b $*( ab+ $*(+ abc $*(+ abc) $*( abc+

$* abc+- $ abc+*

$- abc+*

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Cu trc d liu v gii thut Ng Hu PhcD $- abc+*d/ $-/ abc+*dE $-/ abc+*de# $- abc+*de/

$ abc+*de/-

Thut ton 1

Bc 1 : c mt thnh phn ca biu thc E (dng trung t biu din bng

xu, c t tri qua phi). Gi s thnh phn c c l x

1.1Nu x l ton hng th vit n vo bn phi biu thc E1

1.2Nu x l du ( th y n vo ngn xp

1.3Nu x l mt trong cc php ton +, -, *, / th

1.3.1 Xt phn t y nh ngn xp

1.3.2 Nu Pri(y)>=Pri(x) l mt php ton th loi y ra khi

ngn xp v vit y vo bn phi ca E1 v quay li

bc 1.3.1

1.3.3 Nu Pri(y)


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hin php ton, kt qu c y vo trong ngn xp

Bc 2 : Lp li bc 1 cho n khi ht tt c cc phn t trong biu thc E1.

lc nh ca ngn xp cha gi tr ca biu thc caanf tnh

Bc 3. Kt thc

4.1.5. Bi tp

1. Hy ch ra ni dung ca ngn xp sau mi thao tc trong chui sau

E A S*Y**Q U E***S T***I*ON** . y cc k t ch c ngha

l Push v cc k t * c ngha l Pop.

2. Gii thch cc php PUSH(S, 4), PUSH(S, 1), PUSH(S, 3), POP(S),

PUSH(S, 8), v POP(S) trn cu trc Stack S rng.

3. Hy ci t hai Stck da trn mng mt chiu. Stack c gi l

y khi tng s phn t ca hai Stack bng s phn t ca mng.

4. Vit th tc bng ngn ng Pascal chuyn biu thc t dng

trung t sang dng hu t.

5. Vit th tc tnh gi tr ca biu thc dng hu t

vd : 2 3 4 + *

6. Vit chng trnh i mt s t h thp phn sang h nh phn

s dng cu trc d liu Stack.

4.2. HNG I

4.2.1. Khi nim v hng i

Hng i l mt cu trc d kiu hot ng theo nguyn tc vo trc ra

trc -FIFO.

Trn cu trc ny c hai thao tc c bn l :

1. Thm mi AddQUEUE

2. Loi b -RemoveQUEUE

3. Kim tra hng rng

u vo u ra

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4.2.2. Ci t hng i bng mng:

Ta c th to mt hng i bng cch s dng mt mng 1 chiu vi

kch thc ti a l N (v d, N c th bng 1000) theo kiu xoay vng (coi phn t

an-1 k vi phn t a0).Ta k hiu n l NULLDATA nh nhng phn trc.

Trng thi hng i lc bnh thng:

Q bin hng i, f qun l u hng i, r qun l phn t cui hng i.

Trng thi hng i lc xoay vng (mng rng gia):

Cu hi t ra: khi gi tr f=r cho ta iu g ? Ta thy rng, lc ny hng

i ch c th mt trong hai trng thi l rng hoc y.

Hng i c th c khai bo c th nh sau:

int Q[N];

int F, R;

void init()

{

F=-1;R=-1;

}

void Insert(int *f, int *r, int x)

{

if (*f==-1)

{

*f=0; *r=0;

Q[*r]=x;

}

else

{ if (*f==((*r+1)%N)){

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printf("tran queue");

exit;

}

else

{*r=(*r+1)%N;

Q[*r]=x;

}

}

}int Remove(int *f, int *r)

{ int x;

if (*f==-1) return -1;

x=Q[*f];

if (*f==*r)

{

*f=-1;*r=-1;}

else

*f=(*f+1)%N;

return x;

}

4.2.3. Ci t hng i bng hng danh sch lin kt.

- Khai bostruct Node{

int info;struct Node *next;

a1

an-1

an

.

Front Rear

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Cu trc d liu v gii thut Ng Hu Phc};

Node *F, *R;

- Khi to hng i : F=NULL; R=NULL;

- Thm mt phn t p vo hng i

void Insert(Node **f,Node **r, int x){

Node *p;

p=new Node(); p->info=x; p->next=NULL;

if (*f==NULL)

{ *f=p;*r=p;

}else{

(*r)->next=p;*r=p;

}}

a1

Front

Rear

an-1

a1

an

X

Front

an-2

pRear

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- Loi mt phn t khi hng i

int Remove (Node **f, Node **r)

{Node *p;int x;if (*f==NULL) return 0;

p=*f;x=p->info ;(*f)=(*f)->next ;if (*f==NULL) *r=NULL;

delete p;

return x;

}

4.2.4. Bi ton m phng s dng hng i

Bi ton qun l kho hng :(qun l phiu nhp hng, qun l phiu xut, qun

l hng tn kho, lng tin tn). qun tin tn, hng tn th ngi ta thng s

dng hng i hng nhp trc s c bn trc.M phng my ch phc v cc yu cu t my gi n s dng cc dch v

khc nhau. M phng trung tm thng tin phc v khch hng. M phng truyn tin

gia hai my tnh.

4.2.5. Bi tp

1. Xy dng mt hng i m cc phn t l cc nhn vin trong mt cng ty.

Yu cu : Nhn vin : h tn, tui, qu quan. Tm kim nhn vin theo hoten,

tuoi, quequan.

2. Dng cc php ton c bn ca hng i v ngn xp, vit mt thut ton

o ngc cc phn t ca hng i.

3. M phng trung tm thng tin phc v khch hng.

4.3. Hng i u tin

Hng u tin l mt kiu d liu tru tng trn tp hp cng vi cc php

ton : thm (INSERT) v xo phn t b nht (DELETEMIN), MAKENULL khito.

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Mi phn t a trong hng i c u tin l p(a).

Php ton DELETEMIN s cho ra kt qu l phn t c u tin nh nht

trong tp hp v xa n khi tp hp ny. Nh vy ta c th coi DELETEMIN l t

hp ca th tc DELETE v hm MIN tho lun trong chng.

Thut ng hng gi cho chng ta hnh dung cc phn t ca tp hp ang

xp hng ch phc v, thut ng u tin gi cho ta hiu rng khng phi l " n

trc- c phc v trc".

V d ti bnh vin, cc bnh nhn xp hng ch phc v nhng khng phi

ngi n trc th c phc v trc m h c u tin theo tnh trng khn cp

ca bnh.

C th ci t hng i u tin bng danh sch lin kt hoc cu trc cy.

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Bi 5. CC THUT TON SP XP.5.1. Bi ton sp xp.

Sp xp l qu trnh b tr li cc phn t ca mt tp hp no (hoc cc

mu tin) theo mt th t xc nh. V d th t tng dn (gim dn) i vi mt dy

s, theo th t t in i vi mt dy k t,

5.2. Sp xp chn.

tng ca thut ton chn trc tip m phng mt trong nhng cch sp xp

t nhin nht trong thc t:

Chn phn t nh nht trong N phn t ban u, a phn t ny v v tr ng

l u dy hin hnh; sau khng quan tm n n na, xem dy hin hnh ch cn

N-1 phn t ca dy ban u, bt u t v tr th 2; lp li qu trnh trn cho dyhin hnh... n khi dy hin hnh ch cn 1 phn t. Dy ban u c N phn t, vy

tm tt tng thut ton l thc hin N-1 lt vic a phn t nh nht trong dy

hin hnh v v tr ng u dy. Cc bc tin hnh nh sau : Bc 1 : i = 1; Bc 2 : Tm phn t a[min] nh nht trong dy hin hnh t a[i] n a[N] Bc 3 : Hon v a[min] v a[i] Bc 4 : Nu i


38/64

Cu trc d liu v gii thut Ng Hu Phc nh gi gii thut

i vi gii thut chn trc tip, c th thy rng bc th i, bao gi cng

cn (n-i) php so snh xc nh phn t nh nht hin hnh. S lng php so

snh ny khng ph thuc vo tnh trng ca dy s ban u, do vy trong mitrng hp c th kt lun :

S ln so snh =

S ln hon v (mt hon v bng 3 php gn) li ph thuc vo tnh trng ban

u ca dy s, ta ch c th c lc trong tng trng hp nh sau :

Trng hp S ln so snh S php gn

Tt nht n(n-1)/2 0Xu nht n(n-1)/2 3n(n-1)/2

5.3. Sp xp chn.

Gii thut

Gi s c mt dy a1 , a2 ,... ,an trong iphn t u tin a1 , a2 ,... ,ai-1 c

th t. tng chnh ca gii thut sp xp bng phng php chn trc tip l tm

cch chn phn t ai vo v tr thch hp ca on c sp c dy mi a1 , a2

,... ,ai tr nn c th t. V tr ny chnh l v tr gia hai phn t ak-1v ak tha ak-1 = 0) and (a[pos] > x)) do

Begina[pos+1] := a[pos];

pos:=pos -1;end;a[pos+1] := x;

end;End;

Ci tin thut ton : khi tm v tr thch hp chn a[i] vo on a[0] n

a[i-1], do on c sp, nn c th s dng gii thut tm nh phn thc hin

vic tm v tr pos, khi c gii thut sp xp chn nh phn :

nh gi gii thut

i vi gii thut chn trc tip, cc php so snh xy ra trong

mi vng lp while tm v tr thch hp pos, v mi ln xc nh v tr

ang xt khng thch hp, s di ch phn t a[pos] tng ng. Giithut thc hin tt c N-1 vng lp while , do s lng php so snh

v di ch ny ph thuc vo tnh trng ca dy s ban u, nn ch

c th c lc trong tng trng hp nh sau :

Trng hp S php so snh S php gn

Tt nht

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Xu nht

5.4. Sp xp ni bt.

Gii thut

tng chnh ca gii thut l xut pht t cui (u) dy, i ch cc cp

phn t k cn a phn t nh (ln) hn trong cp phn t v v tr ng u

(cui) dy hin hnh, sau s khng xt n n bc tip theo, do vy ln x l

th i s c v tr u dy l i. Lp li x l trn cho n khi khng cn cp phn t

no xt. Cc bc tin hnh nh sau : Bc 1 : i = 1; // ln x l u tin

Bc 2 : j = N; //Duyt t cui dy ngc v v tr i

Trong khi (j < i) thc hin:

Nu a[j]N-1: Ht dy. Dng

Ngc li : Lp li Bc 2.

V d

Cho dy s a: 12 2 8 5 1 6 4 15

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Ci t

Procedure BubleSort(var a:m1c; N:integer );var i, j,tg:integer;Begin

for i:= 2 to N-1 dofor j :=N-1 downto i do

if (a[j]< a[j-1]) thenbegin tg:=a[j];a[j]:=a[j-1];a[j-1]:=tg; end;

End;

nh gi gii thut

i vi gii thut ni bt, s lng cc php so snh xy ra

khng ph thuc vo tnh trng ca dy s ban u, nhng s lng

php hon v thc hin ty thuc vo kt qa so snh, c th c lc

trong tng trng hp nh sau:

Trng hp S ln so snh S ln hon v

Tt nht 0

Xu nht

5.5. Sp xp nhanh.

sp xp dy a1, a2, ..., an gii thut QuickSort da trn vic phn hoch dy

ban u thnh hai phn:

Dy con 1:Gm cc phn t a1.. ai c gi tr khng ln hn x

Dy con 2:Gm cc phn t ai .. an c gi tr khng nh hn x

vi x l gi tr ca mt phn t ty trong dy ban u. Sau khi thc hin

phn hoch, dy ban u c phn thnh 3 phn:

1. ak< x , vi k = 1..i

2. ak= x , vi k = i..j

3. ak> x , vi k = j..N

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Trong dy con th 2 c th t, nu cc dy con 1 v 3 ch c 1 phn t

th chng cng c th t, khi dy ban u c sp. Ngc li, nu cc dy

con 1 v 3 c nhiu hn 1 phn t th dy ban u ch c th t khi cc dy con 1, 3

c sp. sp xp dy con 1 v 3, ta ln lt tin hnh vic phn hoch tng dy

con theo cng phng php phn hoch dy ban u va trnh by .

Ci t

void QuickSort(int *a, int trai, int phai)

{

int i,j;int x;

x = a[(trai+phai)/2]; // ch?n ph?n t? gi?a lm gi tr? m?c

i =trai; j = phai;

do {

while(a[i] < x) i++;

while(a[j] > x) j--;

if(i


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nhng n gin, d din t gii thut, phn t c v tr gia thng

c chn, khi k = (l +r)/ 2

Gi tr mc x c chn s c tc ng n hiu qu thc hin thut ton v

n quyt nh s ln phn hoch. S ln phn hoch s t nht nu ta chonc x l phn t median ca dy. Tuy nhin do chi ph xc nh phn t

median qu cao nn trong thc t ngi ta khng chn phn t ny m chn

phn t nm chnh gia dy lm mc vi hy vng n c th gn vi gi tr

median

nh gi gii thut

Hiu qa thc hin ca gii thut QuickSort ph thuc vo vic chn gi tr

mc. Trng hp tt nht xy ra nu mi ln phn hoch u chn c phn t

median (phn t ln hn (hay bng) na s phn t, v nh hn (hay bng) na s

phn t cn li) lm mc, khi dy c phn chia thnh 2 phn bng nhau v

cn log2(n) ln phn hoch th sp xp xong. Nhng nu mi ln phn hoch li

chn nhm phn t c gi tr cc i (hay cc tiu) l mc, dy s b phn chia

thnh 2 phn khng u: mt phn ch c 1 phn t, phn cn li gm (n-1) phnt, do vy cn phn hoch n ln mi sp xp xong. Ta c bng tng kt

Trng hp phc tp

Tt nht n*log(n)

Xu nht n2

5.6. Heap Sort.5.6.1. tng:

Nhn xt: Khi tm phn t nh nht bc i, phng php sp xp chn

trc tip khng tn dng c cc thng tin c c do cc php so snh bc i-

1.

V l do trn ngi ta tm cch xy dng mt thut ton sp xp c th

khc phc nhc im ny.

Mu cht gii quyt vn va nu l phi tm ra c mt cu trc

d liu cho php tch ly cc thng tin v s so snh gi tr cc phn t trong qua

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trnh sp xp.

Gi s d liu cn sp xp l dy s : 5 2 6 4 8 1 c b tr theo quan

h so snh v to thnh s dng cy nh sau :

Trong mt phn t mc i chnh l phn t ln trong cp phn t mc

i+1, do phn t mc 0 (nt gc ca cy) lun l phn t ln nht ca dy.Nu loi b phn t gc ra khi cy (ngha l a phn t ln nht v

ng v tr), th vic cp nht cy ch xy ra trn nhng nhnh lin quan n phn t

mi loi b, cn cc nhnh khc c bo ton, ngha l bc k tip c th s dng

li cc kt qu so snh bc hin ti.

5.6.2. nh ngha Heap:

Heap c nh ngha l mt dy cc phn t ap, a2 ,... , aq tho cc quanh sau vi mi i thuc [p, q]:

1/. ai >= a2i

2/.ai >= a2i+1

{(ai , a2i), (ai ,a2i+1) l cc cp phn t lin i }

5.6.3. Tnh cht ca Heap :

Mi dy a1 , a2 ,... , an, dy con aj, aj+1,, an to thnh mt heap vij=(n div 2 +1).

5.6.4. Gii thut Heapsort :

Gii thut Heapsort tri qua 2 giai on :

Giai on 1 :Hiu chnh dy s ban u thnh heap;

Giai on 2: Sp xp dy s da trn heap:

Bc 1: a phn t ln nht v v tr ng cui dy:

r = n; Honv (a1 , ar );

Bc 2: Loi b phn t ln nht ra khi heap: r = r-1;

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Hiu chnh phn cn li ca dy t a1 , a2 ... ar thnh mt heap.

Bc 3: Nu r>1 (heap cn phn t ): Lp li Bc 2

Ngc li : Dng

Da trn tnh cht ca Heap ta c th thc hin giai on 1 bng cch

bt u t heap mc nhin an/2+1 , an/2+2 ... an, sau thm dn cc phn t an/2,

an/2-1, ., a1 ta s nhn c heap theo mong mun.

// Heap sort Mang a bat dau tu 1

void Shift (int *a, int trai, int phai )// hieu chinh

{ int x,i,j;

i = trai; j =2*i; // (ai , aj ), (ai , aj+1) l cc ph?n t? lin ??i

x = a[i];

while (j


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}

void HeapSort (int *a, int N)// N chi so ben phai cung

{ int r;

CreateHeap(a,N);

r = N; // r l v? tr ?ng cho ph?n t? nh? nh?t

while(r > 0)

{

Hoanvi(&a[1],&a[r]);

r = r -1;

Shift(a,1,r);}

}

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Bi 6. CU TRC CY.6.1. Khi nim v cy.

6.1.1. nh ngha:

Mt cy l mt tp hp hu hn cc nt, trong c mt nt c bit gi l nt gc,

gia cc nt c quan h phn cp cha-con.

C th nh ngha cy mt cch qui nh sau:

1/ Mt nt l mt cy v chnh l nt gc ca cy ny.

2/ Nu T1, T2,Tk l cc cy vi cc nt gc l n1, n2,, nk. Gi s n0 l

mt nt c cc nt con l n1, n2,, nk th ta s c cy T vi nt gc l n0.

Qui c cy khng c mt nt no c l cy rng

Cc ng dng ca cy:

+ Biu din cy th mc

+ Biu din cc tp bao nhau

+ Biu din biu thc s hc

6.1.2. Cc khi nim:

Bc ca mt nt : l s con ca nt .

Bc ca mt cy : l bc ln nht ca cc nt trong cy ( s cy con ti a ca

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mt nt trong cy ). Cy c bc n th gi l cy n-phn.

Nt gc : nt khng c nt cha

Nt l : khng c nt con - nt bc 0.

Nt nhnh : l nt c bc khc 0 v khng phi l gc.

Mc ca mt nt :

+ Mc (gc(T))=0.

+ Gi T1,T2,T3,.,Tn l cy con ca T0.

Mc (T1)=Mc(T2)=Mc(T3)=.=Mc(Tn)=Mc(T0)+1.

Chiu cao ca mt nt: L di ng i t nt n nt l xa nht.

su ca mt nt (mc ca mt nt): L chiu di ng i t nt gc nnt .

Chiu cao ca mt cy: L chiu cao ca nt gc.

Nhn xt

Trong cu trc cy khng tn t chu trnh.

T chc mt cu trc cy cho php truy cp nhanh n cc phn t ca

n.

6.1.3. Cy c gc:

nh ngha: Trong mt cy, nu ta chn mt nh c bit gi l gc ca

cy v nh hng cc cnh trn cy c hng t gc i ra th ta c mt th c

hng gi l cy c gc.

Ch : Cng mt cy, nu ta chn gc khc nhau th cy c gc thu c s

khc nhau.V d

a

b

df

g

h

a

a

b b

c

cc

d

d

f

f

g gh h

ee

e

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T1 T2 T3

6.1.4. Cy c th t:

Nu ta phn bit th t cc nt con ca cng mt nt th cy gi l cy c th

t, th t qui c t tri sang phi. Nh vy, nu k th t th hai cy sau l hai cykhc nhau:

Trong trng hp ta khng phn bit r rng th t cc nt th ta gi l cy

khng c th t. Cc nt con cng mt nt cha gi l cc nt anh em rut (siblings).

Quan h "tri sang phi" ca cc anh em rut c th m rng cho hai nt bt k theo

qui tc: nu a, b l hai anh em rut v a bn tri b th cc hu du ca a l "bn tri"

mi hu du ca b.

6.2. Cy nh phn.

6.2.1. Cc khi nim c bn.

nh ngha:

Cy nh phn l mt cu trc cy c c im mi nt c ti a 2 nt con.

Tnh cht:

(1) S lng ti a cc nt mc i trn cy NP l 2i

CM: i=0 => c ti a 1 nt ng

I=1 => c ti a 2 nt ng

Gi s ng vi i-1 => c 2i-1 nt

Cn chng minh ng vi i: mc i -1 c 2 i-1 nt, vi cy NP mi nt c ti a

2 nt con => mc i c ti a 2.2i-1 = 2i (pcm)

(2) S lng ti a cc nt trn cy NP c chiu cao h l 2(h+1)-1 (h>=0)

T (1)=> s nt ti a trn cy c chiu cao h l

20+21+ ...+ 2h=a1* (1-qn)/(1-q)=20*(1-2h+1)/(1-2)=2(h+1)-1

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6.2.2. Biu din cy nh phn.

+) Dng mng biu din cy nh phn

nh s cc nt theo chiu t tri sang phi, t trn xung di ta thy xut

hin qui lut sau: Con ca nt th i l cc nt v tr 2*i v 2*i+1.Cha ca nt j l nt v tr j div 2

=> c th lu tr cy NP bng mng

VD:

Hn ch: vi cy NP khng y mng s rng nhiu.

+) Dng danh sch mc ni

struct NODE

{ int Key;

struct NODE *pLeft, *pRight;

};

NODE *Root;void init(NODE **Root)

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{

*Root=NULL;

}

6.3. Thm cc nt trn cy.

6.3.1. Thm cc nt trn cy theo th t trc (Node-Left-Right).

void NLR(NODE *r)

{ if (r != NULL)

{ printf ("%d--",r->Key );

NLR(r->pLeft);

NLR(r->pRight);

}

}

6.3.2. Thm cc nt trn cy theo th t gia (Left- Node -Right).

void LNR(NODE *r)

{ if (r != NULL)

{ LNR(r->pLeft);printf ("%d--",r->Key );

LNR(r->pRight);

}

}

6.3.3. Thm cc nt trn cy theo th t sau (Left-Right-Node).

void LRN(NODE *r){ if (r != NULL)

{ LRN(r->pLeft);

LRN(r->pRight);

printf ("%d--",r->Key );

}

}

6.4. Bi tp.

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Bi 7. CY NH PHN TM KIM.7.1. Khi nim cy nh phn tm kim.

nh ngha: Cy nh phn tm kim (CNPTK) l cy nh phn trong ti mi nt, kha ca ntang xt ln hn kha ca tt c cc nt thuc cy con tri v nh hn kha ca tt c cc ntthuc cy con phi.

Di y l mt v d v cy nh phn tm kim:

Nh rng buc v kha trn CNPTK, vic tm kim tr nn c nh hng.

Hn na, do cu trc cy vic tm kim tr nn nhanh ng k. Nu s nt trn cy l

N th chi ph tm kim trung bnh ch khong log2N.

Trong thc t, khi xt n cy nh phn ch yu ngi ta xt CNPTK.

7.2. Tm mt phn t x trong cy.

NODE* searchNode(NODE *r, int X)

{

if(r!=NULL){

if(r->Key == X) return r;

if(r->Key > X)

return searchNode(r->pLeft, X);

else

return searchNode(r->pRight, X);

}return NULL;

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}

Kh qui

NODE *searchNode1(NODE *r, int x)

{ NODE *p;

p = r;

while (p != NULL)

{

if(x == p->Key) return p;

else

if(x < p->Key) p = p->pLeft;

else p = p->pRight;}

return NULL;

}

D dng thy rng s ln so snh ti a phi thc hin tm phn t X l h, vi h l chiucao ca cy. Nh vy thao tc tm kim trn CNPTK c n nt tn chi ph trung bnh khongO(log2n) .

V d: Tm phn t 55

7.3. Thm mt phn t x vo cy.

Vic thm mt phn t X vo cy phi bo m iu kin rng buc ca

CNPTK. Ta c th thm vo nhiu ch khc nhau trn cy, tuy nhin thm vo mt

nt l s l d thc hin nht, do ta c th thc hin qu trnh tng t thao tc tmkim. Khi chm dt qu trnh tm kim cng chnh l lc tm c ch cn thm.

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Hm insert tr v gi tr 1, 0, 1 khi khng b nh, gp nt c hay thnh

cng:

int insertNode(NODE **r, int X)

{ if((*r)!=NULL){ if((*r)->Key == X)return 0;

if((*r)->Key > X)

return insertNode(&((*r)->pLeft), X);

else

return insertNode(&((*r)->pRight), X);

}

(*r) = new NODE();

if((*r) == NULL) return -1;

(*r)->Key = X;

(*r)->pLeft =NULL; (*r)->pRight = NULL;

return 1;

}

7.4. Xa nt trn cy.Vic hy mt phn t X ra khi cy phi bo m iu kin rng buc ca

CNPTK.

C 3 trng hp khi hy nt X c th xy ra:

X - nt l.

X - ch c 1 cy con (tri hoc phi).

X c c 2 cy con

Trng hp th nht: ch n gin hy X v n khng mc ni n phn t no khc.

Trng hp th hai: trc khi hy X ta mc ni cha ca X vi con duy nht ca n.

Trng hp cui cng: ta khng th hy trc tip do X c 2 con Ta s hy gin

tip. Thay v hy X, ta s tm mt phn t th mng Y. Phn t ny c ti a mt con.

Thng tin lu ti Y s c chuyn ln lu ti X. Sau , nt b hy tht s s l Y

ging nh 2 trng hp u.

Vn l phi chn Y sao cho khi lu Y vo v tr ca X, cy vn l CNPTK.

C 2 phn t tha mn yu cu:

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Phn t nh nht (tri nht) trn cy con phi.

Phn t ln nht (phi nht) trn cy con tri.

// Remove mot nut

void searchFor(NODE **p, NODE **q)

{

if( (*q)->pLeft!=NULL )

searchFor(p, &((*q)->pLeft) );

else

{ (*p)->Key= (*q)->Key;

*p=*q;// nut can xoa

*q = (*q)->pRight;}

}

int delNode(NODE **r, int X)

{

if((*r)==NULL) return 0;

if((*r)->Key > X)

return delNode ( &((*r)->pLeft), X);if((*r)->Key < X)

return delNode (&((*r)->pRight), X);

else

{ //T->Key == X

NODE *p = *r;

if((*r)->pLeft == NULL)

(*r) = (*r)->pRight;

else if((*r)->pRight == NULL)

(*r) = (*r)->pLeft;

else

{ searchFor(&p, &((*r)->pRight)); }

delete p;

}

}

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7.5. Cy biu thc.

V d : Cy sau y s biu din cho biu thc (a + b) * (a - c)

Trong cy trn th n1, n2,..., n7 l cc tn nt cn *, +, -, a, b, c l cc

nhn ca nt.

Mi nt l s biu din cho mt ton hng n c.

Mi nt trung gian s biu din cho mt ton t. Gi s nt n biu din

cho ton t 2 ngi, nt con bn tri biu din cho biu thc E1, nt con bn

phi biu din cho biu thc E2 th nt n s biu din cho biu thc E E2

Thm cc nt trn cy theo th t trc cho chng ta biu thc tin t (Prefix),

Theo th t gia cho chng ta biu thc trung t (Infix)

V theo th t sau cho chng ta biu thc hu t (Postfix) ca biu thc ton

hc ban u.

V d: i vi cy biu thc c cho v d trn, ta c:

+ Biu thc tin t : * + a b - a c.

+ Biu thc trung t : a + b * a - c.

+ Biu thc hu t : a b + a c - *.

7.6. Cy quyt nh:

Cy biu din li gii ca bi ton thng qua nhiu la chn gi l cy quyt

nh. V d:

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Bi 8. CC PHNG PHP TM KIM8.1. Bi ton tm kim.

Bi ton tm kim c m t nh sau: Hy xc nh v tr ca phn t x trong

mt bng lit k cc phn t phn bit x1, x2, x3, , xn hoc xc nh n khng c

mt trong bng lit k .

Li gii ca bi ton l v tr xut hin ca x trong bng lit k hoc a ra

thng bo l x khng c trong bng.

8.2. Tm kim tun t.

Tm kim tun t l phng php tm kim n gin v t nhin nht, phng

php ny thng c nhiu ngi s dng, nht l trong trng hp tm kim trn

file. Tuy vy tc tm kim ca phng php ny thng rt chm.TTon: TTon bt u bng vic so snh phn t cn xc nh Item vi phn

u tin x1, nu cha thy th so snh vi cc phn t k tip cho n khi tm ra phn

t cho hoc l t n phn t cui cng ca danh sch.

8.3. Tm kim nh phn.

8.3.1. Tm kim nh phn trn cu trc mng

Mun thc hin tm kim nh phn th trc ht chng ta phi sp xp d liuu vo (gi s sp xp tng dn).

TTon:

B1: Phm vi tm kim l ton b danh sch.

B2: So snh phn t cn xc nh Item vi phn t chnh gia ca phm vi

tm kim (gi l x).

Nu: Item> x th phm vi tm kim mi s l cc phn t sau ca x.

Nu: Item< x th phm vi tm kim mi s l cc phn t trc ca x.

Nu: Item=x th thng bo l tm thy, kt thc chng trnh.

B3: Nu tn ti phm vi tm kim th lp li bc 2, ngc li th kt thc

chng trnh v thng bo l khng tm thy.

int Binary_Search(int *a, int n, int Item)

{ int last, first, mid;

bool found;

found=false;

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first=0; last=N-1;

while ((!found)&&(firstItem)

last=mid-1;

else

if (Item>a[mid])

first =mid+1;

else

{ found=true;

return mid;

}

}

return -1;

}

BT1: Lit k cc bc tm kim s 9 trong dy:

1, 3, 4, 5, 6, 8, 9, 11

8.3.2. Tm kim trn cy nh phn (trong phn CNPT)

8.4. Tm kim da vo gi tr kha

8.4.1. Hm bm.

Trong hu ht cc ng dng, kho c dng nh mt phng thc truy

xut d liu mt cch gin tip. Hm bm l hm c dng nh x mt kho vo

mt a ch trn bng bm.

Chng ta gi h(k) l gi tr hm bm ca kha k. Trng hp h(ki) =h(kj) th

hm bm b ng , khi ta phi x l thm sao cho vn lu tr c ia ch

ring bit ca cc bn ghi c kha ki v kj.

8.4.2. Hm bm kiu chia

L hm c tnh ton d trn pha chia ly phn d h(k)= k mod m, m l mts nguyn t.

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Nh vy k s nhn cc gi tr 0, 1, 2,, m-1.Vd: m=7; k=10-> h(k)= 10 mod

7= 3

8.4.3. Hm bm kiu nhn

Gi tr kha c nhn vi chnh n.Vd:

S th t K H(k) Ly 3 ch s u tin

1 11 121

2 111 123

8.5. Cc phng php khc phc ng

C nhiu phng php khc nhau khc phc ng , tuy nhin chung ta c

th chia chng thnh 2 loi:

(1) Phng php a ch m: Da trn nguyn tc nh sau: vi a ch bn

th ngi ta tm cc a ch cn trng ngay tip sau thay th.

(2) Phng php mc xch: Cc bn ghi xy ra ng c lu vo mt danh

sch tuyn tnh.8.5.1. Phng php a ch m

*) Phng php th tuyn tnh

Gi s x l kha m ta cn xc nh a ch trong bng bm.

h0 = h(x) nu h0 bn th ta tm cc a ch tip theo hi=(h0+i) mod m vi

i=1,2,..

Khi n cui bng m vn cha tm c a ch trng th quay v u

bng, tm tip cho n khi thy a ch trng hoc quay v ng a ch ban u =>

bng bm trn.

vd: cho cc kha k[i]=140, 122, 178, 110, 160, 147, 291, 182,300,305

m=7

h(ki) mod 7 =0, 3, 3, 5,6,0,4,0,6,4

0 140

1 1472 182

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3 122

4 178

5 110

6 1607 291

8 300

9 305

Nxt:

Bng bm ny ch ti u khi bm u, ngha l, trn bng bm cc khi ccha vi phn t v cc khi cha s dng xen k nhau, tc truy xut lc ny c

bc 0(1). Trng hp xu nht l bm khng u hoc bng bm y, lc ny hnh

thnh mt khi c c n phn t, nn tc truy xut lc ny c bc 0(n).

*) Phng php cu phng

h0=h(x) = x mod m;

Nu h0bn th cc a ch tip theo s c xt: hi=( h0+i) mod m;i =1,2,.., m

div 2.Nhc im:

C th xy ra trng hp c (m div 2) a ch th u bn => bng bm trn,

tuy nhin l trong bng bm vn cn a ch trng.

8.5.2. Phng php mc xch

*) Phng php mc ni ngoi

#define M 100struct node

{

int key;

struct node *next

};

//khai bao kieu con tro chi nut

Node * hashTable[M]

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*) Phng php mc ni trong

Ch dng khng gian a ch c ca bng bm (khng cp php thm nh

trong phng php mc ni ngoi).

Thm kha x: nu h(x) cn trng th thm vo a ch ny v n s l a ch

u tin ca danh sch mc ni.

Nu h(x) bn th xy ra 2 kh nng:

(gi k l kha hin din ti a ch h(x))

- Nu h(k)=h(x) => kha x s thuc danh sch lin kt c nt u tinti h(k). Tm kha x trong danh sch ny: Nu thy php tm kim

tha mn. Nu khng thy x th phi tm mt a ch trng (d tun t)

thm kha x vo v b sung x vo danh sch lin kt.

- Nu h(k)h(x) => Ta phi tm trng a kha k sang . Ch

phi iu chnh cc con tr ti n v xut pht t n. a kha x vo

v tr h(x).

8.6. ng dng ca hm bm- Trong bo mt thng tin: M ha RSA, ch k s

PP m ho RSA:

Chn 2 s nguyn t p,q ln, N=p*q;

Chn s nguyn t E (1


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P=5; Q=11; chn E=7; Tnh D theo cng thc

E*D=1 (mod (p1-)*(q-1))= 1 (mod 40)

7 * D = 1 (mod 40)

7 * D = K * 40 + 1, K l mt s no

Chn K=4 -> D=23

7 * 23 = 161 = 4 * 40 + 1

E l kha cng khai dng m ha, D l kha ring (b mt) dng gii m

Tin hnh m ha, gii m hai s 31 v 7

M ha: 31^7 (mod 55) = 27512614111 (mod 55) =26

7^7 (mod 55) = 823543 (mod 55) =28

Gii m:26^23 (mod 55) =

350257144982200575261531309080576 (mod 55) =31

28^23 (mod 55) =

1925904380037276068854119113162752 (mod 55) = 7

Ch k s:

Gi s c mt cng ty bun bn qu cho php nhn n hng qua Internet,

c cc yu cu sau:+Ch cng ty mi c kh nng nhn n hng. (khch hng mun n hng

ca h c gi b mt)

+Cng ty c th kim tra chnh xc mt n hng c phi ca khch hng X

hay khng.

Cch thc hin:

Mi khch hng X chn mt eX lm m cng khai, dX m b mt.

Cng ty cng chn eC v dC

Vi yu cu 1: X s m ha H bng eC Cty gii m bng dC

Vi yu cu th 2: X s m ha bng dX, sau li m tip bng eC Cty

gii m bng dC, sau bng eX.

- S dng hm bm to cc file ch s trong h qun tr CSDL SQL Server v

Oracle.

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Bi 9. Cy cn bng

Bi 10. Mt s thut ton nn

10.1. Run Length

10.2. Huffman

10.3. LZ77,LZ78

10.4. LZW

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