Network flow [HDU 1565] grid fetch (1)

Check count (1)

Time limit:10000/5000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 5961 Accepted Submission (s): 2268

Problem description gives you a n*n checkered checkerboard with a non-negative number in each lattice.
Take out a number of the number, so that any of the two number of the lattice does not have a common edge, that is, the number of the 2 lattice is not adjacent, and the number of the largest.

Input includes multiple test instances, each of which consists of an integer n and n*n non-negative numbers (N<=20)

Output for each test instance, the maximum possible and

Sample Input375 15 21 75 15 28 34 70 5

The Output188 of the maximum point weight independent set of the Sample for the EK algorithm

/************************************************author: Hart 13Created TIME:2015/1/7 23:38:02file Name:hdu 1565 squares Fetch number (1). cpp*************************************************/#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>#include<vector>#include<queue>#include<Set>#include<string>#include<cmath>#include<cstdlib>#include<ctime>using namespacestd;#defineINF 0x3f3f3f3f#definell Long Long#defineN 510intN;intsrc;intdes;intsum;intPre[n];intMpt[n][n];intMap[n][n];intdir[4][2]={1,0,-1,0,0,1,0,-1};queue<int>Q;BOOLBFs () { while(!q.empty ()) Q.pop (); memset (PRE,-1,sizeof(pre)); PRE[SRC]=0;    Q.push (SRC);  while(!Q.empty ()) {        intu=Q.front ();        Q.pop ();  for(intv=0; v<=n*n+1; v++)        {            if(pre[v]==-1&& mpt[u][v]>0) {Pre[v]=u; if(V==des)return 1;            Q.push (v); }        }    }    return 0;}intEK () {intmaxflow=0;  while(BFS ()) {intminflow=INF;  for(intI=des;i!=src;i=Pre[i]) Minflow=min (minflow,mpt[pre[i]][i]); Maxflow+=Minflow;  for(intI=des;i!=src;i=Pre[i]) {Mpt[pre[i]][i]-=Minflow; Mpt[i][pre[i]]+=Minflow; }    }    returnMaxflow;}intMain () {inti,j,k;  while(SCANF ("%d", &n)! =EOF) {Sum=0; memset (MPT,0,sizeof(MPT));  for(i=1; i<=n;i++)        {             for(j=1; j<=n;j++) {scanf ("%d",&Map[i][j]); Sum+=Map[i][j]; }} src=0; Des=n*n+1;  for(i=1; i<=n;i++)        {             for(j=1; j<=n;j++)            {                if((i+j)%2==0) mpt[src][(I-1) *n+j]=Map[i][j]; Elsempt[(I-1) *n+j][des]=Map[i][j]; }        }         for(i=1; i<=n;i++)        {             for(j=1; j<=n;j++)            {                if((i+j)%2==0)                {                     for(k=0;k<4; k++)                    {                        intx=i+dir[k][0]; inty=j+dir[k][1]; if(x>=1&& x<=n && y>=1&& y<=N) {mpt[(i-1) *n+j][(x1) *n+y]=INF; } }}}} printf ("%d\n", sum-EK ()); }    return 0;}

Network stream [HDU 1565] Box fetch number (1)