baitap_thuchanh(ltdt)

8/4/2019 baitap_thuchanh(ltdt)

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BI TP V THC HNH

MN HC

L thuyt th

2009


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MC LC

CHNG 1: I CNG V TH .............................................................31Xt v d thc t.......................................................................................................................3

2Cc thut ng th................................................................................................................4

3Biu din cc th v s ng cu th ............................................................................5

4Tnh lin thng.........................................................................................................................6

CHNG 2: TH EULER V TH HAMILTON ................................9CHNG 3 TH C TRNG S V NG I NGN NHT................................13

...................................................................................................................................................14

CHNG 4: CY ................................................................................................17

Vit tiu lun............................................................................................................................24

TI 1:..............................................................................................................................24

S DNG PHNG PHP TH TH HIN VIC B TR LCH THI CHO SINHVIN KHOA TON TIN HC. .........................................................................................24

TI 2:..............................................................................................................................25

S DNG PHNG PHP TH GII BI TON DN GIAN (BI TON 1).....25

TI 3: .............................................................................................................................25

S DNG PHNG PHP TH GII BI TON DN GIAN (BI TON 2).....25

TI 4: .............................................................................................................................26

CC THUT TON TM NG I NGN NHT TRN TH..................................26

TI 5: CC GII THUT TM CY PH TI TIU ..................................................27

TI 6: BI TON T CHC THI CNG.....................................................................27

TI 7: BI TON QUN L D N.............................................................................28

TI 8: BI TON 8 QUN HU...............................................................................28 TI 9: BI TON QUN M I TUN...................................................................29

TI 10: ...........................................................................................................................30

BI TON PHN B KNH TRUYN HNH TI BSCL.................................................30

DNG THUT TON T MU TH.............................................................................30

TI 11: ...........................................................................................................................30

BI TON PHN B KNH TRUYN HNH TI MIN NG NAM B.......................30

DNG THUT TON T MU TH.............................................................................30

TI 12: ...........................................................................................................................31

1


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Bi tp v thc hnh mn l thuyt th

CHNG 1: I CNG V TH

1 Xt v d thc t

Bi 1.1. Vi mi trng hp sau, v cc m hnh th biu din cc ng bay v nir v loi th c dng. Trong lch bay mi ngy nh sau:

- T TP.HCM: c mt chuyn n H Ni, mt chuyn n Nng, mt chuyn nPh Quc, mt chuyn n Ngh An, mt chuyn n Hi Phng;

- T H Ni: c hai chuyn n TP.HCM, mt chuyn n Nng, mt chuyn nNgh An, mt chuyn n Hi Phng;

- T Nng: c mt chuyn n Hi Phng, hai chuyn bay n TP.HCM; mt chuynn H Ni;

- T Ngh An: c mt chuyn n H Ni, mt chuyn n TP.HCM;

- T Hi Phng: c mt chuyn n H Ni, mt chuyn n TP.HCM, v mt chuyn

n Nng;- T Ph Quc: c mt chuyn n TP.HCM.

a) th biu din cc thnh ph c chuyn bay gia chng.

b) th biu din s chuyn bay hot ng gia cc thnh ph, cng vi mt khuynbiu th chuyn du lch c bit ngm cnh thnh ph, ct v h cnh ti Ph Quc.

c) th biu din y thng tin v hng bay v s chuyn bay gia cc thnh ph.

Phn hng dn

a) th v hng

b) a th v hng

c) a th c hng

3

Ph QucNgh An

H Ni

Hi Phng Nng

TP. HCM

Ph QucNgh An

H Ni

Hi Phng Nng

TP. HCM


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Bi 1.2. Xc nh xem th no sau y l th n, a th, th c hng.

Bi 1.3. Trong trn u vng trn, i H thng i G, i C, v i A; i G thng i Av i C; i C thng i A. Hy m hnh ha kt qu ny bng mt th c hng

2 Cc thut ng th

Bi 2.1. Xc nh s lng cc nh, s lng cc cnh, v bc ca cc nh trong cc th sau. Cho bit nh no l nh c lp, nh no l nh treo.

4

Ph Quc Ngh An

H Ni

Hi Phng Nng

TP. HCM

a)

b)

c)d)

ca

f e d

a)

b a

e d cb)

b

b

h g ec)

c d

if

a


6/34

a

b

c

d

e

f f b

da)

ba

dc

e

a f b

e

dgc

b)

a

c d

b

a

d c

b

Bi tp v thc hnh mn l thuyt thBi 2.2. Tm tng cc bc ca cc nh trong cc th cc Bi 2.1, v kim chngrng n bng hai ln s cc cnh trong th.

Bi 2.3. C th tn ti mt th n c 15 nh, mi nh c bc bng 5 khng? Tisao?

Bi 2.4. Trong mt bui chiu i, mi ngi u bt tay nhau. Chng t rng tng sngi c bt tay l mt s chn. Gi s khng ai t bt tay mnh.

Bi 2.5. Xc nh s nh, s cnh, s bc vo v s bc ra ca mi nh i vi th chng sau.

Bi 2.6. Hy xc nh tng cc bc vo v tng cc bc ra cc nh ca th trong bi2.5 mt cch trc tip. Chng t rng chng u bng tng cc cnh ca th.

Bi 2.7. th s c bao nhiu cnh nu n c cc nh bc 4, 3, 3, 2, 2. V mt thnh vy.

Bi 2.8. C tn ti th n cha nm nh vi cc bc sau y? Nu c hy v th.

a) 3, 3, 3, 3, 2. b) 1, 2, 3, 4, 5.

c) 1, 2, 3, 4, 4.

Bi 2.9. V tt c cc th con ca th sau.

Bi 2.10. Tm hp ca cc cp th n sau

3 Biu din cc th v s ng cu thBi 3.1. Dng danh sch k biu din cc th sau.

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Bi 3.2. Biu din cc th trong bi 3.1 bng ma trn k.

Bi 3.3. V cc th ng vi ma trn k c cho nh sau.

a)

0 1 0

1 0 1

0 1 0

b)

0 0 1 1

0 0 1 0

1 1 0 1

1 1 1 0

c)

1 1 1 0

0 0 1 0

1 0 1 0

1 1 1 0

Bi 3.4. Dng ma trn lin kt biu din cc th trong Bi 3.1.Bi 3.5. Xc nh xem cc cp th cho c l ng cu khng.

4 Tnh lin thng

Bi 4.1. Cc danh sch nh sau y c to nn ng i trong th bn di haykhng? ng i no l n?ng i no l chu trnh? di ca cc ng i ny l

bao nhiu?

a) (a, e, b, c, b)

b) (a, e, a, d, b, c, a)

c) (e, b, a, d, b, e)d) (c, b, d, a, e, c)

6

a

c d

b

a)

a

d c

b

b)

u4

u5

u2

u3

v1

v2

v4

v5

u1

v3

a)


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Bi tp v thc hnh mn l thuyt thBi 4.2. Cc danh sch nh sau y c to nn ng i trong th bn di haykhng? ng i no l n? ng i no l chu trnh? di ca cc ng i ny l

bao nhiu?

a) (a, b, e, c, b)

b) (a, d, a, d, a)

c) (a, d, b, e, a)

d) (a, b, e, c, b, d, a)

Bi 4.3. Xc nh xem cc th cho c lin thng khng.

Bi 4.4. C bao nhiu thnh phn lin thng trong cc th cc Bi tp 4.3? Tm ccthnh phn lin thng .

Bi 4.5. Tm tt c cc nh ct v cnh ct ca th.

Bi thc hnh s 1: Biu din thBi tp 1:

Nhp vo ma trn k ca mt n th (t bn phm v c t tp tin).

a. Kim tra tnh hp l ca th (gi tr trn ng cho chnh u bng 0).b. Kim tra xem th l v hng hay hu hng?c. Nu ma trn k c nhp t bn phm th xut ra thnh tp tin matranke.txtd. Nu ma trn c c t tp tin th xut kt qu ma trn ra mn hnh hin th.e. Xut ra bc ca tt c cc nh ca th (s cnh ni ti nh).

f. Kim tra tnh lin thng ca th? Xut ra tt c cc thnh phn lin thng nu c.Bi tp 2:

Nhp vo ma trn trng s ca mt n th (t bn phm v c t tp tin).

a. Kim tra tnh hp l ca th (gi tr trn ng cho chnh u bng 0).b. Kim tra xem th l v hng hay hu hng?c. Nu ma trn k c nhp t bn phm th xut ra thnh tp tin trongso.txtd. Nu ma trn c c t tp tin th xut kt qu ma trn ra mn hnh hin th.e. Xut ra cnh c trng s nh nht v ln nht.

Hng dn:

Chng trnh nhp vo ma trn k ca th t bn phm

#include

#include

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Bi tp v thc hnh mn l thuyt thmain()

{

int n,m,i,j;

int a[100][100];

// Doc n, m tu ban phim

printf(" Nhap n, m ");scanf("%d %d",&n,&m);

// Doc mang a kich thuoc n*m

for (i=0;i


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Bi tp v thc hnh mn l thuyt thfor (j=0;j


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Bi 4. Xc nh s tn ti chu trnh Euler trong cc th c hng sau. V cc chu trnhny nu chng tn ti.

Bi 5.Xc nh xem th c hng trong Bi 5.4 c ng i Euler hay khng. V ccng i Euler ny nu c.

Bi 6. Xc nh th cho c cha chu trnh Hamilton hay khng. Nu c hy tm mtchu trnh nh th. Nu khng c hy gii thch l do v sao khng tn ti.

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Bi tp v thc hnh mn l thuyt thBi 7 th trong Bi 5.6 c ng i Hamilton khng? Nu c, hy tm ng . Nukhng c, cho bit l do ti sao khng tn ti mt ng i nh vy.

Bi thc hnh s 2: Duyt thBi tp 1:

Ci t thut ton duyt theo chiu su v chiu rng vi th c cho bi ma trn k c ttp tin matranke.txt, chng trnh cho php nhp vo nh xut pht v hin th kt qu duyt.

Bi tp 2:

Ci t chng trnh cho php nhp vo 2 nh x v y ca th, kim tra xem c ng i t xti y (v ngc li) hay khng?

Hng dn:

4.1.1 Duyt th theo chiu su (thut ton DFS)

tng chnh ca thut ton c th trnh by nh sau: Ta s bt u tm kim t mt nh v0 no ca th. Sau chn u l mt nh tu k vi v 0 v cha c duyt. Lp li qu trnh i vi u. y l th tc quy. Qu trnh quy kt thc khi khng chn c nh u na.

Th tc quy c vit nh sau:

Th tc DFS(v);

(*tim kiem theo chieu sau bat dau tu dinh v;

cac bien Chuaxet, Ke la bien toan cuc*)Bt u

Tham_dinh(v);

Chuaxet[v]:=false;

For u l Ke(v) do

If Chuaxet[u] then DFS(u);

Kt thc (*dinh v da duyet xong*)

Duyt ton b th theo chiu su:

Bt u

for v thuc V do Chuaxet[v]:=true;

for v thuc V do

if Chuaxet[v] then DFS(v);

Kt thc

phc tp l : O(n+m)

4.1.2 Duyt th theo chiu rng (thut ton BFS)

tng chnh ca thut ton nh sau:

S dng 1 hng i lu tr cc nh s c duyt. Ban u a nh bt u duyt u vo hng i. Thc hin qu trnh lp cho n khi hng i rng. Mi bc lp lm nh sau:

o Ly 1 nh v ra khi hng i. Thc hin cc cng vic cn thit vi nh .o Xc nh cc nh w k vi nh v m cha duyt.o a cc nh w ny vo hng i.

Th tc c vit nh sau:Th tc BFS(u);

(*Tim kiem theo chieu rong bat dau tu dinh u,cac bien Chuaxet, Ke la bien cuc bo*)

Duyt ton b th theo thut ton BFS:

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Bt u

QUEUE:= ;

QUEUE u; (* nap u vao QUEUE*)

Chuaxet[v]:=false;

While QUEUE do

Begin

v QUEUE:; (*lay p tu QUEUE:*)

Tham_dinh(v);

For w l Ke(v) do

If Chuaxet[w] them

Begin

QUEUE w;

Chuaxet[w]:=false;End;

End;

Kt thc

Bt u

(*khi to ban u *)

for mi k thuc V do

Chuaxet[k]:=true;

for u thu V do

if Chuaxet[u] then BFS(v);Kt thc

Thut ton BFS c phc tp: O(n+m)

4.1.3 ng dng ca thut ton duyt

Ta thy, t tng ca thut ton duyt l xut pht t 1 nh u, v i thm cc nh v cn litrong th nu nh tn ti 1 ng i t nh u n v.

Nh vy ta s gii quyt c 2 bi ton nh sau:

Tm 1 ng i gia 2 nh bt k. Kim tra tnh lin thng ca th v tm cc thnh phn lin thng trong th.

xc nh c 1 ng i t nh u n nh v, ta cn phi lm nh sau:

Dng th tc duyt theo chiu su vi nh u. Trong qu trnh duyt c s dng mng cc bin trng thi Chuaxet[k] (hay Free[k])

kim tra nh k c duyt cha. xc nh ng i ta dng mng bin lu tr vt Truoc[k] = t, c ngha l t nh t s

chuyn n nh k.

B sung vo cc th tc DFS v BFS nh sau:

i vi DFS b sung code nh sau:

If Chuaxet[u] then

Begin

Truoc[u]:= v;

Chuaxet[u] = false;

DFS(u);End;

i vi BFS b sung code nh sau:

If Chuaxet [u] then

Begin

QUEUE u;

Chuaxet[u]:= false;

Truoc[u]:= p;End;

Xc nh ng i t u n v nh sau:

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While (v u) do

Begin

In gi tr v ra ngoi mn hnh;

v = Truoc[v] ; // gn v bng nh m t n nhy n v hin ti

End;

In gi tr u ra mn hnh;

xc nh s thnh phn lin thng trong th th ta dng thm 1 bin count m. Khi cc bc phi lm nh sau:

Khi to cc mng bin cn thit, v count = 0. Xt vi mi nh u trong th, nu nh nh u cha c duyt th:

o Tng gi tr ca bin count ln 1 n vo Thc hin cc thut ton duyt i vi nh u.

Nu kt qu bin count = 1 th th lin thng. Trong trng hp cn li th gi tr cacount chnh l s thnh phn lin thng ca th.

Vit thm code xc nh s thnh phn lin thng:

Count :=0;

If Chuaxet[u] then

Begin

Count := Coutn + 1;

DFS(u);

End;

Count :=0;

If Chuaxet[u] then

Begin

Count := Coutn + 1;

BFS(u);

End;

CHNG 3

TH C TRNG S V NG I NGN NHTBi 1. Tm chiu di ca ng i ngn nht gia a v z trong th c trng s sau y

Bi 2. Tm di ca ng ingn nht gia cc cp nh sauy trong cc th c trng s Bi 6.1.

a) a v d b) a v f c) c v

f d) b v zBi 3. Tm di ng i ngnnht t nh A n tt cc nh cn

li trong cc th sau:

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Bi 4. Tm di ng i ngn nht t nh A n tt cc nh cn li trong cc thsau:

Bi 5. Cho th c trng s nh hnh di y. Hy tm ng i ngn nht t nh An nh N.

Bi thc hnh s 3:Bi tp:

Ci t thut ton Ford-Bellman, Dijkstra, Floyd- Warshall vi th c cho bi ma trn trngs c t tp tin trongso.txt, chng trnh cho php nhp vo nh xut pht v hin th kt quduyt.

Hng dn:

4.1.1 Thut ton tm ng i ngn nht Ford-Bellman

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Bi tp v thc hnh mn l thuyt thThut ton ny Ford-Bellman xc nh tt c cc ng i ngn nht t mt nh u chotrc n cc nh v cn li trong th.

T tng thut ton nh sau:

S dng mng D[v] l gi tr khong cch b nht i t nh u n nh v. Ban u ch cD[u] = 0, cn tt c cc nh v cn li u c D[v] = .

ti u cc gi tr D[v] th ta cn chn cc nh trung gian k, sau so snh gi trD[v] hin ti vi tng khong cch D[k] v A[k][v]. Lu y A[k][v] chnh l trngs trn cnh (kv).

Nh vy vi mi nh trung gian k ta cn phi ti u (n-1) nh cn li. Hn na ta c ncch chn gi tr ca k. Nh vy ta c 2 vng lp lng nhau.

Gi s thi im T1, chng ta dng nh k1 l nh trung gian, v ti u c gi tr tinh k2. n thi im T2 , chng ta chn k2 l nh trung gian th li c th ti u cgi tr ti nh k1. Nh vy vic chn 1 nh l trung gian ph thuc vo (n-1) nh cnli. vt ht cc kh nng, ta cng cho mi nh c chn lm trung gian (n-1) ln.

Nh vy ta c vng lp th 3 nm ngoi 2 vng lp trn.

Tuy nhin hiu qu hn, ti mi bc ca vng lp ngoi cng ta kim tra xem cc gitr ti cc nh c thay i khng,nu ti u v khng thay i na th ta s dngchng trnh.

V c bn thut ton m t nh sau:

Procedure Ford_Bellman (u)

Begin

for i:=1 to n-1 do // s ln chn mt nh lm trung gianfor k thuc V\{u} do // chn 1 nh l trung gian

for v thuc V do // ti u vi tt c cc nh cn li

if d[v] > d[k] +a[k][v] then // nu k ti u xy ra

begin

d[v]:=d[k]+a[k][v]; // gn gi tr ti u mi

Truoc[v]:= k; // lu li vt di chuyn.

end;

End;

Mt s ch :

Thut ton trn lm vic vi th c trng s khng m,hoc khng c chu trnh mtng trng s l m.

Thut ton trn lm vic vi ma trn k s c phc tp l O(n3).

4.1.2 Thut ton tm ng i ngn nht Dijkstra

Thut ton ny xc nh ng i ngn nht gia 2 nh c th t u n v.

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Bi tp v thc hnh mn l thuyt thT tng ca thut ton nh sau:

Dng mt tp S lu tr cc nh c duyt. Cch n gin nht phn bit gia nh duyt vi nh cha duyt l dng mng bin trng thi Free[k].

Dng mng ph D[k] xc nh gi tr khong cch ngn nht i t u n k. Ban u ta chgn D[u] = 0 v D[k] = vi v u.

Bt u qu trnh duyt cc nh k khc u v qu trnh lp ch kt thc khi :o t n nh ch v .o Hoc khng th duyt tip c na.

Mi bc ca qu trnh duyt nh sau:o Xc nh nh k trong s cc nh cha c duyt, sao cho D[k] l b nht.o Nu k c chn chnh l nh ch v ta kt thc vng lp.o Nu gi tr D[k] c chn l (tc l khng chn thm c na) ta cng kt

thc vng lp.o Trong trng hp cn li, ta a k vo tp S (gn Free[k] = 0) v ti u cc nh

cn li theo nguyn tc sau: nu D[t] > D[k] + A[k][t] D[t] = D[k] + A[k][t];

Thut ton c m t nh sau:

Procedure Dijstra (u, v);

Begin

while True do

begin

Tm nh k tho mn Free[k] = 1 v D[k] nh nht ;

if D[k] = hoc k = v then BREAK;For v thuc V do

If Free[v] = 1 v d[v]>d[k]+a[k][v] then

begin

d[v]:=d[k]+a[k][v];

Truoc[v]:=u;

end;

end;

End;

Mt s ch :

Thut ton lm vic vi th c trng s khng m,hoc khng c chu trnh m tngtrng s m.

phc tp ca thut ton l O(n2).

4.1.3 Thut ton tm ng i ngn nht Floyd- Warshall

Thut ton ny xc nh ng i ngn nht gia mi cp nh trong th.

T tng thut ton nh sau:

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Bi tp v thc hnh mn l thuyt th xc nh ng i ngn nht gia mi cp nh trong th, th ta phi xc nh tt c

n*(n-1) gi tr, tng ng vi cc cp (u,v) trn th. Khi ta dng lun ma trn trng s Anxn lm ma trn lu kt qu. Khi A[i][j] l gi

tr khong cch ngn nht i t nh i n nh j. T tng thut ton ny cng kh n gin:

o Chn 1 nh l trung gian, gi s l nh k.o Ti u ha ng i t nh u n nh v theo k. Tc l ta so snh A[u][v] vi

tng A[u][k] v A[k][v]. Nu i t u n v di hn so vi i t u qua k ri n vth ta s ti u gi tr A[u][v].o Ta thy c n cch chn nh k, vi mi k c n cch chn nh xut pht u v vi

mi u ta c n cch chn nh kt thc v. Nh vy ta c 3 vng lp lng nhau.

Cng ng dng t tng ny xc nh cc thnh phn lin thng ca th . y l thutton Warshall. T tng chnh l:

Nu t u k c ng i, tc l A[u][k] = 1 v t k n v cng c ng i, tc l A[k][v] = 1 th chc chn c ng i t u n v A[u][v] = 1.

C 2 thut ton u c phc tp l : O(n3).Thut ton c m t nh sau:

Procedure Floyd_Warshall

Begin

For k = 1 to n do

For u = 1 to n do

For v = 1 to n do

If A[u][v] > A[u][k] + A[k][v] then

A[u][v] = A[u][k] + A[k][v];

End;

Procedure Warshall

Begin

For k = 1 to n do

For u = 1 to n do

For v = 1 to n do

If A[u][k] = 1 v A[k][v] = 1 then

A[u][v] = 1;

End;

CHNG 4: CYBi 1. Tm cy khung nh nht bng thut ton Prim ca th gm cc nh A, B, C, D,E, F, H, I c cho bi ma trn trng s sau.

.

Bi 2. Tm cy khung nh nht ca th sau theo thut ton Kruskal v Prim.

17

A B C D E F H

A

I

B

C

D

E

F

H

I

15 16 19 23 20 32 18

15 33 13 34 19 20 12

16 33 13 29 21 20 19

19 13 13 22 30 21 11

23 34 29 22 34 23 21

20 19 21 30 34 17 18

32 20 20 21 23 17 14

18 12 19 11 21 18 14


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a42

104


Bi 3. Tm cy khung nh nht bng thut ton Prim ca th gm cc nh A, B, C, D,E, F, H, I c cho bi ma trn trng s sau.

Yu cu vit cc kt qu trung gian trong tng bc lp, kt qu cui cng cn a ra tpcnh v di ca cy khung nh nht.

Bi thc hnh s 4: Cc thut ton v cy

Bi tp: Ci t cc thut ton tm cy khung nh nht.

Hng dn:

R rng 1 th cho ta nhiu cy khung. Vn l xc nh cy khung no trong th c trngs sao cho tng trng s l b nht.

Thut ton Prim Dijsktra xc nh cy khung b nht.

T tng ca thut ton ny l da trn thut ton tm ng i ti u Dijsktra.

Dng tp S lu tr cc nh c duyt. ban u S = . i vi bi ton tm ng i ngn nht gia 2 im u,v, th thut ton Dijsktra s ln

lt chn cc nh trn th sao cho gi tr khong cch t nh u n nh mi

duyt l nh nht c th c. Qu trnh dng li khi nh v c duyt. i vi bi ton xc nh cy khung nh nht, th qu trnh duyt cc nh cng nh

trn, nhng ch dng li khi khng cn duyt c thm nh no na. Ch y bin D[v] khng phi l khong cch i t nh u n v m l khong

cch t v n S (l khong cch ngn nht t v n 1 nh trong tp S).

18

16 15 23 19 18 32 20

16 13 33 24 20 19 11

15 13 13 29 21 20 19

23 33 13 22 30 21 12

19 24 29 22 34 23 21

18 20 21 30 34 17 14

32 19 20 21 23 17 18

20 11 19 12 21 14 18

A B C D E F G H

AB

C

D

E

F

GH


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Bi tp v thc hnh mn l thuyt tho Nu gi tr khong cch ti nh cui cng , th ta c 1 cy khung b nht.o Trong trng hp ngc li th khng xc nh c cy khung b nht.

Thut ton c m t nh sau: phc tp ca thut ton l O(n2).

Procedur Prim_Dijsktra;

Begin

Chn 1 nh u l gc ca cy. D[u] = 0; D[v] = vi v u; S = while True do

begin

Tm nh k tho mn Free[k] = 1 v D[k] nh nht ;

If (khng xc nh c k OR D[k] = ) then Break.

Else F := F U (T[k],k), S = S U k ; Free[k] = 0

For v V do

IfFree[v] = 1 v D[v]> A[k][v] then // ch k ny!!!!!

begin

D[v]:= A[k][v]; T[v]:=k;

end;

end;

End;

4.1.4 Thut ton Kruskal (thut ton tham lam n tham)

T tng ca thut ton nh sau: Khi to tp cc cnh ca cy khung F = . Chn cc cnh ca th theo trng s t nh n ln, sao cho n khng to ra chu trnh

trong tp F. a cnh va chn vo tp F v xa n khi tp cnh E ca th. Qu trnh lp li cho n khi trong tp F c ng n-1 cnh.

Thut ton m t nh sau:

Procedure Kruskal;

Begin

F := ; // khi to tp rng

While |F| < (n-1) and (E ) do // kim tra s phn t ca tp F

Begin

Chn cnh {e} sao cho trng s trn {e} l nh nht.

E:=E\{e};

if (F U {e} khng cha chu trnh) then F := T U { e} ;

End;

if (|T| < n-1) then th khng lin thng;

End;

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phc tp ca thut ton l O(m*log2m) vi |E| = m.

Bi tp tng hp:1. Cho ma trn k (trng s) ca mt th nh sau

a. V th tng ng

b. th c phi l th Euler (na Euler) hay khng? Gii thch.Nu c,tm chu trnh (ng i) Euler trong th.c. th c phi l th Hamilton hay khng? Nu c, tm chutrnh Hamilton trong th.d. Tm ng i ngn nht t nh s 1 n nh s 8.e. Tm cy khung nh nht ca th.

2. Cho ma trn k (trng s) ca mt n th v hng nh sau:

a. V th, cho bit bc ca cc nh. y c phi l th Eulerhay khng?b. Tm ng i ngn nht t nh 1 n cc nh cn li bng thutton Dijkstra.c. Tm cy khung nh nht ca th bng thut ton Kruskal.

17. Cho th nh hnh v:

a. Xc nh bn bc ra v bn bc vo ca cc nh ca th.b. y c phi l th lin thng mnh khng? Ti sao?c. Xy dng ma trn k, trng s ca th.

d. Dng thut ton Dijkstra tm ng i ngn nht t nh 1 n nh3.18. Cho th nh hnh v:

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a. Xc nh bc ca cc nh ca th, t cho bit y c l thEuler hay na Euler hay khng?b. Xy dng ma trn k trng s ca thc. Dng thut ton Dijkstra tm ng i ngn nht t nh 1 n nh

5 trong th.d. Tm cy khung nh nht ca th.

Bi tp nng cao:

1. Bi ton truyn tin trn mng.

Mt mng my tnh gm N my nh s t 1 n N, v M knh truyn tin mt chiu gamt s cp my trong mng c nh s t 1 n M. Mng my tnh l thng sut,ngha l t mt my c th truyn tin n mt my bt k khc bng cc ng ni trctip hoc thng qua cc my trung gian. Mt my trong mng l s chn nu s knhtruyn tin trc tip t n n cc my khc l chn. Mt my trong mng l s l nu sknh truyn tin trc tip t n n cc my khc l l. Gi s s v t l hai my l trongmng, hy i hng truyn tin ca mt s knh bin i mng cho thnh mng(khng nht thit phi thng sut) m trong n hia my s v t tr thnh 2 my chn vkhng lm thay i tnh chn l ca cc my khc trong mng. S knh i hng cng tcng tt.

D liu nhp vo t file vn bn Net.inp:

- Dng u tin cha 2 s N, M (N


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Bi tp v thc hnh mn l thuyt thhin tip yu cu khc. Cn lp k hoch cho thang my thc hin M yu cu ri trv tng mt sao cho tng qung ng thang my phi di chuyn l t nht.

D liu nhp vo t file vn bn Tm.inp

- Dng u ghi 2 s nguyn dng N v M.

- M dng tip theo, dng th I ghi 2 s ai v bi m t yu cu th i

Kt qu ghi ra file vn bn Tm.out

- Dng u l tng qung ng tm c

- Dng th 2 m t th t thc hin cc cng vic bi mt hon v ca 1, 2,M.

3. Bi ton K ng i ngn nht

Vng t X c N thnh ph (3


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4. Bi ton cng cha kn chng

Magachip IV the Splendid, vua ca Byteotia, c nh chn ph m cho cng cha Adaca mnh. Cng cha th mun chng mnh phi l ngi thng minh, khng keo kit vkhng lng ph qu. Nn nh vua suy ngh mi chn phng php th ti bng cch chnlu i ca mnh lm ni thi u. Lu i gm nhiu phng trng by qu him v nivi nhau bng dy cc hnh lang thn dn c th tham quan, phng cui cng l phngca cng cha, thm mt phng phi tr mt ng bytealer v s xut pht t phngu tin. Nh vua trao cho mi chng trai n cu hn mt s ng tin cha trong 1 citi v yu cu l mi ngi thm 1 s phng sao cho n phng cui cng th tiu ht stin cho, nu n phng cui cng m vn cn tin th phi thm li mt s phng

trng by na mi quay li. Hy lp trnh gii quyt vn trn 1 chng trai lun c

th thc hin c yu cu ca nh vua.

D liu nhp t file Zam.inp miu t lu i, s hiu phng cng cha

ang , tng s tin trong ti:

- Dng u tin c 5 s nguyn dng n(s lng phng), m(shnh lang), e(s hiu phng xut pht), p(s hiu phng cngcha), b(tng s tin vua ban cho).

- Dng 2 c n s nguyn dng ci, mi s l chi ph mi ln vothm trong phng i.

- Trong m dng tip theo c tng cp s nguyn dng (x, y),mi cp ni biu th mt hnh lang

ni phng x vi phng y.

Tnh ra dy cc phng i qua n phngca cng cha v lu hnh trnh vo fileZam.out

V d: (nh bng d liu bn cnh)

5. Bi ton chia ct ch

Cui nm 1944, qun Lin X phn cng

vo vng X, ni c N thnh ph b phtxt c chim ng. Dc theo cc conng gia 2 thnh ph u c qun ccanh gi. B tham mu qun s ca LinX mi ch o phn cng tiu dit ch,

bc u thc hin chia ct qun chthnh 2 vng tch bit chng khnglin lc c vi nhau, nhng cn tnhton tiu dit nh th no l li nht mch cn huy ng lc lng t nht ginh thng li. Bit rng i tiu dit qun

c th Lin X phi b tr s qun tnht bng vi s qun ch chim ng.Hy vit chng trnh gip b tham muqun Lin X ch cn iu t nht lclng ginh chin thng.

23

Zam.inp

5 6 3 4 91 2 3 4 5

2 4

5 4

1 5

1 2

2 3

3 1

Zam.out

3 2 4

Chiacat.inp Chiacat.out

10 18

1 2 8

1 4 4

1 7 1

2 3 5

2 4 5

2 9 93 4 4

3 6 2

3 9 7

4 5 5

5 6 6

5 7 5

5 8 3

6 8 7

6 10 1

7 8 5

8 10 1

9 10 9

10

1 7

3 6

4 5

6 10

8 10


25/34


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Bi tp v thc hnh mn l thuyt thNgn ng s dng : C, C++, Visual C++, Visual Basic

TI 2:

S DNG PHNG PHP TH GII BI TON DN

GIAN (BI TON 1).

C T TI :

C mt v khch n xin nh vua ban cho mt qu cam trong vn ng uyn. Nhvua chp thun. ng ta n vn mi hay phi qua 3 cng lnh canh.

n cng th nht, ngi lnh canh bo v khch: " Vua ban cho th anh c vo mhi, nhng lc ra phi a cho ta mt na s cam v thm mt tri".

Qua cng th hai v th ba, hai lnh canh cng ni vi anh nh ngi lnh canh

th nht. V khch phi hi bao nhiu qu cam lc ra khi vn cn c mt qutrong tay?

YU CU CA TI

V l thuyt : Tm hiu v trnh by cc khi nin c bn v :

- Cc th tc (hm) c lin quan n giao din ca mn hnh ha.

- th v cc khi nim c bn v th c hng, th v hng.

- Minh ha bi ton bng th.

- Thit lp thut ton.

- C th m rng bi ton bng cch cho ngi s dng thay i kt qu s lngcam m v khch c c sau khi ra khi vn.

V lp trnh:

- Vit chng trnh da vo thut ton thit lp.

- Giao din thn thin vi ngi s dng.

- Kt qu cho ra l mt th vi mu sc phn bit.

MI TRNG CI T

Ngn ng s dng : C, C++, Visual C++, Visual Basic

TI 3:

S DNG PHNG PHP TH GII BI TON DN

GIAN (BI TON 2).C T TI :C hai cha con ngi nng dn i mua v tu ha. Ngi bn v tu hi: " Ch b

ny bao nhiu tui ?". ng cha tr li: "Con trai ti tui gp 5 ln em gi n, m n tui

25


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Bi tp v thc hnh mn l thuyt thgp 6 ln tui n. Tui ti th bng tui ca v v hai con ti cng li. Cn m ti th

bng tui ca tt c gia nh chng ti cng li". Ngi bn v tu ni: "Thi ri! Conng c min v".

Da vo u m ngi v tu min v cho ch b? Bit rng, theo lut ng stth tr em di 6 tui i cng ngi ln s c min v.

YU CU CA TI

V l thuyt : Tm hiu v trnh by cc khi nin c bn v :

- Cc th tc (hm) c lin quan n giao din ca mn hnh ha.

- th v cc khi nim c bn v th c hng, th v hng.

- Minh ha bi ton bng th.

- Thit lp thut ton.

V lp trnh:

- Vit chng trnh da vo thut ton thit lp.- Giao din thn thin vi ngi s dng.

- Kt qu cho ra l mt th vi mu sc phn bit.

MI TRNG CI T

Ngn ng s dng : C, C++, Visual C++, Visual Basic

TI 4:

CC THUT TON TM NG I NGN NHT TRN TH

C T TI :

Vn dng cc l thuyt c bn v th ci t chng trnh cho php biu din th, kim tra tnh lin thng v tm ng i ngn nht gia 2 nh cho trc bng gii thutDijkstra, Ford-Bellman trn th v hng.

YU CU CA TI :

L thuyt: Cc thao tc c bn v ha. Cc khi nim v th c hng v th v hng Cc cch biu din th, cc phng php tm kim trn th (tm

theo chiu rng v chiu su) v tnh lin thng. Cc gii thut c lin quan nh: kim tra tnh lin thng, tm ng i

ngn nht. Nhng cu trc d liu cn thit ci t chng trnh.

Chng trnh:Phi c nhng chc nng c bn sau:

Cp nht d liu v th. Biu din th trn mn hnh. Kim tra tnh lin thng.

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Bi tp v thc hnh mn l thuyt th Cho php tm ng i ngn nht gia 2 nh bt k.

MI TRNG CI T :

Ngn ng lp trnh s dng: C hay C ++

TI 5: CC GII THUT TM CY PH TI TIU

C T TI :

Vn dng cc l thuyt c bn v th ci t chng trnh cho php biu din th, kim tra tnh lin thng v tm cy c trng lng nh nht bng gii thut Kruscal.

YU CU CA TI :

L thuyt:

Cc thao tc c bn v ha. Cc khi nim v th c hng v th v hng Cc cch biu din th, cc phng php tm kim trn th (tm

theo chiu rng v chiu su) v tnh lin thng. Cc gii thut c lin quan nh: kim tra tnh lin thng, gii thut

kim tra tnh lin thng v gii thut Kruscal tm cy c trng lngnh nht.

Nhng cu trc d liu cn thit ci t chng trnh. Chng trnh:

Phi c nhng chc nng c bn sau:

Cp nht d liu v th. Biu din th trn mn hnh. Kim tra tnh lin thng. Cho php tm cy c trng lng nh nht.

MI TRNG CI T :


TI 6: BI TON T CHC THI CNG

C T TI :

Vn dng cc l thuyt c bn v th ci t chng trnh cho php biu din th, biu din th sau khi xp hng, xc nh cc thi im sm nht, tr nht ca tng cngvic, thi gian hon thnh cng trnh v v s GANT th hin k hoch hon thnh cngtrnh.

YU CU CA TI :

L thuyt: Cc thao tc c bn v ha. Cc khi nim v th c hng v th v hng Cc cch biu din th, Cc php biu din th.

27


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Bi tp v thc hnh mn l thuyt th Gii thut xp hng trn th, gii thut xc nh cc thi gian sm

nht v thi gian tr nht. Nhng cu trc d liu cn thit ci t chng trnh.


Cp nht d liu v bi ton t chc thi cng. Biu din th trc v sau khi xp hng ln mn hnh.

Xc nh cc thi im sm nht, tr nht ca tng cng vic, thi gianhon thnh cng trnh

V s GANT

MI TRNG CI T :


TI 7: BI TON QUN L D N

C T TI :

Vn dng cc l thuyt c bn v th ci t chng trnh cho php biu din th, biu din th sau khi xp hng, xc nh cc thi im sm nht, tr nht ca tng cngvic, thi gian hon thnh cng trnh v v s GANT th hin k hoch hon thnh cngtrnh.

YU CU CA TI :

L thuyt: Cc thao tc c bn v ha. Cc khi nim v th c hng v th v hng Cc cch biu din th, Cc php biu din th. Gii thut xp hng trn th, gii thut xc nh cc thi gian sm

nht v thi gian tr nht. Nhng cu trc d liu cn thit ci t chng trnh.


Cp nht d liu v bi ton t chc thi cng. Biu din th trc v sau khi xp hng ln mn hnh. Xc nh cc thi im sm nht, tr nht ca tng cng vic, thi gian

hon thnh cng trnh V s GANT

MI TRNG CI T :

S dng: MS . PROJECT 2003

TI 8: BI TON 8 QUN HU

C T TI :

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Bi ton : t 8 qun hu trn bn c vua 8 8 sao cho khng c qun hu no c thtn cng c con khc (theo lut chi c vua), ngha l phi t cc qun hu sao chokhng c hng, ct hoc ng cho no trn bn c c hn 1 qun hu. Chng hn, mtcch t qun hu ng nh sau :

QQ

QQ

QQ

QQ

YU CU CA TI :

V l thuyt :- Nm vng l thuyt c bn v cu trc d liu v gii thut- Gii thut tm kim su kt hp quay lui (Backtracking)V lp trnh:- Ci t cu trc d liu t chc bn c.- Ci t thut ton tm kim su kt hp quay lui theo nguyn tc : Trc tin, t qunhu vo th nht ca ct 1, r rng tt c cc ca ct b khng ch nn khngth t qun hu khc. t tip mt qun hu vo ct th hai, hai u ca ct bcm bi qun hu th nht, do vy ta t qun hu vo th ba. Tip tc vi ct th ba, u tin c th t qun hu ca ct ny l th nm Tip tc vi cc ct cn li trn

bn c cho n khi tm c mt li gii ng.- Hin th bn c sau mi nc i.- Dch chng trnh sang file thc thi.

Ngn ng lp trnh s dng : C, C ++, Visual C++

TI 9: BI TON QUN M I TUNC T TI :

Bi ton : Trn bn c vua 8 8, mt qun m c php i theo lut c vua. V tr u

tin ca qun m t ti mt no . Hy tm cch di chuyn qun m qua tt c cc ca bn c sao cho mi ch c i qua 1 ln duy nht. Chng hn 10 v tr hp l utin cho qun m nu qun m bt u khi hnh ti (1, 1) trn bn c vua nh sau :

1 4 62 5 7

38

9

... 10

YU CU CA TI :

29


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Bi tp v thc hnh mn l thuyt thV l thuyt :- Nm vng l thuyt c bn v cu trc d liu v gii thut.- Thut ton quiV chng trnh:- Ci t cu trc d liu t chc bn c.- Khi to ngu nhin v tr t qun m u tin.- Ci t chng trnh my tnh qui theo kiu th sai, vt cn mi kh nng tm li

gii: tm kim nc i k tip bng cch chn mt trong nhng c th t qun m hpl tip theo trn bn c. C tip tc cho nhng nc sau n khi tm thy mt li gii.- Hin th bn c sau mi nc i.- Dch chng trnh sang file thc thi.Ngn ng lp trnh s dng : C, C ++, Visual C++

TI 10:

BI TON PHN B KNH TRUYN HNH TI BSCL

DNG THUT TON T MU THC T TI :

Bi ton (Phn b knh truyn hnh ng bng Sng Cu long): Cc knh truynhnh t s 2 n s 13 c phn chia cho cc i truyn hnh cc tnh trong vng ng

bng Sng Cu long (C Mau, Bc Liu, Sc Trng, Cn th, Vnh long, An giang, Kingiang, ng thp, Tr Vinh, Bn Tre, Tin giang, Long An) sao cho khng c hai i

pht no hai tnh nm cnh nhau trn bn a l li dng cng mt knh ?

YU CU CA TI :

V l thuyt:- Nm vng l thuyt c bn v cu trc d liu v gii thut- Gii thut t mu th (gio trnh Ton ri rc, Nguyn c Ngha, NXB H QucGia TP. H Ch Minh)V chng trnh :- D liu nhp vo v xut ra di dng file cha cc thng tin v th.- T chc th: Mi i pht tng ng 1 nh, mi mu biu th 1 knh. Vic phn chiaknh tng ng vi vic t mu th (t mu theo thut ton Greedy xt ln lt theos th t cc nh)

- Ci t qu trnh thc hin gii thut t mu (hin th Tn tnh v S knh tng ng)- Dch chng trnh sang file thc thi.

Ngn ng lp trnh s dng : C, C ++ , Visual C++

TI 11:

BI TON PHN B KNH TRUYN HNH TI MINNG NAM B

DNG THUT TON T MU THC T TI :

Bi ton (Phn b knh truyn hnh min ng Nam B): Cc knh truyn hnh t s15 n s 25 c phn chia cho cc i truyn hnh cc tnh trong vng min ng Nam

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Bi tp v thc hnh mn l thuyt thB (TP.H Ch Minh, ng Nai, B Ra Vng Tu, Ty Ninh, Bnh Dng, BnhPhc) sao cho khng c hai i pht no hai tnh nm cnh nhau trn bn a l lidng cng mt knh ?

YU CU CA TI :

V l thuyt:

- Nm vng l thuyt c bn v cu trc d liu v gii thut- Gii thut t mu th (gio trnh Ton ri rc, Nguyn c Ngha, NXB H QucGia TP. H Ch Minh)V chng trnh :- D liu nhp vo v xut ra di dng file cha cc thng tin v th.- T chc th: Mi i pht tng ng 1 nh, mi mu biu th 1 knh. Vic phn chiaknh tng ng vi vic t mu th (t mu theo thut ton Greedy xt ln lt theos th t cc nh)- Ci t qu trnh thc hin gii thut t mu (hin th Tn tnh v S knh tng ng)- Dch chng trnh sang file thc thi.

Ngn ng lp trnh s dng : C, C ++ , Visual C++

TI 12:

XY DNG BN TRC TUYN NHNG CON NGLN TRONG NI THNH TP.H CH MINH VI VICTNH CHI PH THP NHT DI CHUYN GIA TRUNGTM CC QUN.

C T TI :Vn dng cc l thuyt c bn v th ci t chng trnh cho php xc nh con

ng i trn th, kim tra tnh lin thng v tm ng i ngn nht gia 2 nh , mi nh l1 a danh c th ti trung tm Tp.H Ch Minh bng gii thut Dijkstra, Ford-Bellman trn th v hng.

YU CU CA TI :

L thuyt: Cc thao tc c bn v ha. Cc khi nim v th c hng v th v hng

Cc cch biu din th, cc phng php tm kim trn th (tmtheo chiu rng v chiu su) v tnh lin thng. Cc gii thut c lin quan nh: kim tra tnh lin thng, tm ng i

ngn nht. Nhng cu trc d liu cn thit ci t chng trnh.


Cp nht d liu v th. Biu din th trn mn hnh. Kim tra tnh lin thng. Cho php tm ng i ngn nht gia 2 nh bt k.

MI TRNG CI T :


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34/34


TI 15: BI TON NG I NGI A TH

C T TI

Ni dung bi ton:

Xt bi ton v ngi a th, xut pht t Bu in Tp. H Ch Minh, anh ta in tt cc bu in chnh trong ni , mi ni ng 1 ln. Cui cng anh ta quay tr li

ni xut pht sao cho qung ng m anh ta i qua l ngn nht. Hy tm mt chu trnh(mt ng i khp kn tha mn iu kin trn) sao cho tng di cc cnh l nhnht.

YU CU CA TINm vng c s l thuyt v cu trc d liu. Cc k thut thit k gii thut.Chng trnh cn c cc chc nng sau: Cho nhp vo bi ton: s bu in chnh ti cc

qun, khong cch ni gia cc bu in (c th ly s liu t trong bn hoc kho st thct). Xut ra phng n tm c. Nu th hin di dng ho cng tt.

MT TRNG CI T

Ngn ng lp trnh s dng: C, C++ , Visual C++

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