lời giải tham khảo hsg qg tin học 2014

8/10/2019 Li Gii Tham Kho HSG QG Tin Hc 2014

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LI GII THAM KHOKTHI HSG QG TIN HC 2014

Vit v su tm bi:

Trn Anh Hng Thi Huy v Nguyn Duy Khnh-The VOS Admins-

Bi 1: Con ng Tng Trc[bi](LI GII: Nguyn Duy Khnh)

Ta lu 2*n im vo mng T:

T[i].d l khong cch n vtr bt u ca con

ng

T[i].k l 1 nu l cy tng, l 2 n

u l cy trc

Sp xp mng T theo thttng dn T[i].d

Vi mi T[i], ta phi tm vtr X sao cho tT[i] n

T[x] c t nht a cy tng v b cy trc. t F[i]=X.

Ddng nhn thy: F[i]


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ngha:

F[y] = max(T[x,y]).

Gi ma=max(F[y]) ( vi y


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Sp xp cc im theo gi trx/y.

Cc im c cng gi trchnh l cc im trn

cng mt ng thng. Xl v in kt qu.

O(N^2 log N)

Bi 5: QoS[bi]Thng nht:

+oo : gi trrt ln, dng v cc.

[u] : nh thu

c[u][u] : chi ph cnh t[u] n [u],nu khng

c cnh th t c[u][u] = +oo;

minT[u] : chi ph t nht i t[u] n [T], nu

khng c ng ith t minT[u] = +oo;

Cmin, Tmin : c trong bi.

Thng thng i vi mt bi vthttin, vic

cn lm trc tin lun phi l qun l c scu

hnh (trng hp tha mn).

Trc tin phi m c c bao nhiu ng i t[1]

n [T] m di khng qu Tmin+Cmin.

c nh th, ta sdng phng php qui hoch ng.V cn thttin tb n ln, nn ta sxl

ngc t[T] v[1].

Gi F[u,L]l scch i t[u] n [T] vi i

khng qu L, ta tnh hm nh sau:

Nu minT[u]>L:

F[u,L] = 0 (cc bn tchng minh);Ngc li:

F[u,L] = Sum(F[u,L-c[u][u]]);Ta c thgii thch cng thc nh sau:ti [u] ta cn

qung ng khng qu L, khi sang [u] tn mt
http://vn.spoj.com/problems/QOS/http://vn.spoj.com/problems/QOS/http://vn.spoj.com/problems/QOS/http://vn.spoj.com/problems/QOS/http://vn.spoj.com/problems/QOS/


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c[u][u], nn qung ng cn li n [T] phi khng

c di qu L-c[u][u].

Ta chtnh F[u,L] vi L>=minT[u], nn chch phn

hiu L-minT[u]=P, vy ta i cng thc thnh: E[u,P] lscch i t[u] n [T] vi di khng qu

P+minT[u].

E[u,P] = Sum(E[u,P+minT[u]-minT[u]-c[u][u]]);(Khi i cng thc sthun tin vic tit kim bnh)

Sau khi tnh xong bng gi trE[][], ta gii nh cc

bi thttin khc. phc tp thi gian tnh

ton l O(N^2*Cmin)

Bi 6: JOBSET [bi]u tin ta nhn xt c thchn c mt tp ti

u, ta lm theo 2 bc:

Bc 1: chn vi dn c li nhun dng (Pi>0)

Bc 2: thm vo cc dn cn thit tha mn

iu kin bi.

Bi ny c thtm thy trn mng vi tn ProjectSelection hay tm thy trong phn ng dng ca Max-

flow min-cut theoremtrn trang Wikipedia.

Li gii c vit li theo ti liu ny.

Bi ny gii bng cch a vbi ton lt ct nhnht

trn thc hng gia 2 nh ngun (gi S) v ch

(gi T).

Xy dng thc nh-cnh nh sau:

1.

nh: Gm N nh i din cho N dn v 2 nh S&T

(tng cng: N+2 nh).

2.

Cnh c hng:

Mi Pi > 0, thm mt cnh [S-->i] chi ph ct

Pi;
http://vn.spoj.com/problems/JOBSET/http://vn.spoj.com/problems/JOBSET/http://vn.spoj.com/problems/JOBSET/http://en.wikipedia.org/wiki/Max-flow_min-cut_theoremhttp://en.wikipedia.org/wiki/Max-flow_min-cut_theoremhttp://www.wikipedia.org/http://www.cs.uiuc.edu/class/sp08/cs473/Lectures/lec15.pdfhttp://www.cs.uiuc.edu/class/sp08/cs473/Lectures/lec15.pdfhttp://www.wikipedia.org/http://en.wikipedia.org/wiki/Max-flow_min-cut_theoremhttp://en.wikipedia.org/wiki/Max-flow_min-cut_theoremhttp://vn.spoj.com/problems/JOBSET/


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Mi Pi T] chi ph ct

|Pi| = -Pi;

Mi mi quan hc i phi c j, thm mt cnh

[i-->j] chi ph ct C+1 (vi C l li nhun ti

a c thc, dthy n chnh l tng cc dn c li: C = Sum(Pi) vi Pi>0);

Mi tp dn A tha mn u tng ng vi mt lt ct.

Mi lt ct chia ththnh 2 phn:

1.

Cc nh n c tS: chnh l cc dn c

chn.

2.

Cc nh cn li: cc dn khng c chn.

Ta chng minh c:

e(A) = C - profit(A) profit(A): gi trtp dn A;

e(A): gi trlt ct tng ng;

C: li nhun ti a t c, C=Sum(Pi) vi Pi>0;

profit(A) t ln nht, e(A) cn l lt ct nh

nht.

Bi ton qui vtm lt ct {S,T} nhnht bng thut

ton tm lung cc i.

Theo mnh, nhn ra cc bi ton ng dng lt ct

khng phi l vic ddng. V vy vic chun bkin

thc y trc cc k thi l lun cn thit.

Mrng, ving dng ca bi ton lt ct cc tiu.
http://www.cs.uiuc.edu/class/sp08/cs473/Lectures/lec15.pdfhttp://www.cs.uiuc.edu/class/sp08/cs473/Lectures/lec15.pdfhttp://www.cs.uiuc.edu/class/sp08/cs473/Lectures/lec15.pdfhttp://www.cs.uiuc.edu/class/sp08/cs473/Lectures/lec15.pdfhttp://www.cs.uiuc.edu/class/sp08/cs473/Lectures/lec15.pdf

lời giải tham khảo hsg qg tin học 2014

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