BZOJ3996[TJOI2015] Linear algebra

Description

Give a n*n matrix B and a 1*n matrix C. Find a 1*n 01 matrix A to make

D= (a*b-c) *a^t max. Where A^t is the transpose of a. Output dinput The first line input an integer n, the next n line input B matrix, the first row of the J number represents bij. The next line is to enter n integers, representing the matrix C. Each number in matrix B and Matrix C is a non-negative integer that does not exceed 1000. Output

Output the largest D

Sample Input3
1 2 1
3 1 0
1 2 3
2 3 7Sample Output2HINT

1<=n<=500

Exercises

This is a good packing. ...

D=a*b*a^t-c*a^t

If clause I of a is 1, then D subtracts c[1,i].

If Item I and item J of a are all 1, then D will add b[i,j].

In this way, the title becomes a model similar to BZOJ1497[NOI2006 's maximum profit: N items, taking an item will have a price, and if you take two items at the same time, it will be profitable.

The problem becomes the maximal weight closed sub-graph, and the minimum cut solution is obtained by Dinic network flow.

Code:

Constinf=100000000;typeRec=RecordS,e,w,flow,next:longint; End;varb,bb,d,q:Array[0..2002002] ofLongint; A:Array[-1..2002000] ofRec; V:Array[0..2002000] ofBoolean; N,m,i,j,k,l,st,ed,ww,top,tar,ans,x:longint;functionmin (aa,bb:longint): Longint;begin  ifAa<bb Thenexit (AA); exit (BB);End;procedureAdd (st,ed,ww:longint);beginInc (top); A[TOP].S:=St; A[TOP].E:=Ed; A[TOP].W:=ww; A[top].next:=B[st]; B[ST]:=top;End;functionBfs:boolean;varHead,tail,x,u:longint; Y:rec;beginFillchar (v,sizeof (v), false); Tail:=1; head:=0; d[st]:=1; V[ST]:=true; q[1]:=St;  whileHead<tail Do  beginInc (head); x:=Q[head]; U:=B[x];  whileU>0  Do    beginy:=A[u]; if( notV[Y.E]) and(Y.FLOW&LT;Y.W) Then      beginV[Y.E]:=true; D[Y.E]:=d[x]+1; Inc (tail); Q[tail]:=y.e; End; U:=Y.next; End; End; Exit (V[tar]);End;functionAddflow (p,maxflow:longint): Longint;varO:longint; Y:rec;begin  if(P=tar)or(maxflow=0) Thenexit (Maxflow); Addflow:=0;  whileBb[p]>0  Do  beginy:=A[bb[p]]; if(d[y.e]=d[p]+1) and(Y.FLOW&LT;Y.W) Then    begino:=addflow (Y.e,min (maxflow,y.w-y.flow)); ifO>0  Then      beginInc (A[BB[P]].FLOW,O); Dec (a[bb[p] XOR1].flow,o);        Dec (maxflow,o);        Inc (ADDFLOW,O); ifmaxflow=0  ThenBreak ; End; End; BB[P]:=Y.next; End;End;functionNetwork:longint;beginNetwork:=0;  whileBFs Do  begin     forI:=st toTar Dobb[i]:=B[i];  Inc (Network,addflow (St,inf)); End;End;beginREADLN (n); ST:=0; Top:=1; tar:=N;  fori:=1  toN Do   forj:=1  toN Do  beginread (x); ifX>0  Then    beginInc (TAR);      Add (st,tar,x); Add (Tar,st,0);      Add (Tar,i,inf); Add (I,tar,0);      Add (Tar,j,inf); Add (J,tar,0); Ans:=ans+x; End; End;  Inc (TAR);  fori:=1  toN Do  beginRead (j);    Add (I,TAR,J); Add (Tar,i,0); End; Writeln (ans-network);End.

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BZOJ3996[TJOI2015] Linear algebra