HDOJ 1507-Uncle Tom's Inherited Land *

Question:

Place a rectangle of 1*2 in a matrix... some points cannot be placed in a rectangle. How many rectangles can be placed at most ....

Question:

My first response to this question is status compression DP .. but check the range. okay .. bipartite Graph Matching .. but there is a problem .. if point-to-point side is directly implemented .. there will be confusion .. and it does not conform to the basic model of the 2-Chart (the points on the same side do not have any direct relationship )... so do the bipartite graph .. the first thing is to divide the points into two heaps that will not be directly affected internally .. in this example, (x + y) is an odd or even number .. the rest is bare ..

Program:

#include<iostream>#include<stdio.h>#include<algorithm>#include<cmath>#include<stack>#include<queue>#define ll long long#define MAXN 105using namespace std;int n,match[MAXN],hash[MAXN][MAXN],w[MAXN*MAXN][2];bool arc[MAXN][MAXN],used[MAXN],s[MAXN][MAXN]; bool dfs(int x){       int i;       for (i=1;i<=n;i++)          if (arc[x][i] && !used[i])          {                 used[i]=true;                 if (!match[i] || dfs(match[i]))                 {                       match[i]=x;                       return true;                 }          }              return false; }int getmax(){       int sum=0;       memset(match,0,sizeof(match));       for (int i=1;i<=n;i++)       {               memset(used,false,sizeof(used));               sum+=dfs(i);       }       return sum;}int main(){       int i,j,x,y,m,num,cases=0;         while (~scanf("%d%d",&n,&m) && n)       {                 memset(s,true,sizeof(s));                scanf("%d",&num);                while (num--) { scanf("%d%d",&x,&y),s[x][y]=false; };                num=0;                memset(hash,0,sizeof(hash));                for (x=1;x<=n;x++)                  for (y=1;y<=m;y++)                     if (s[x][y]) hash[x][y]=++num,w[num][0]=x,w[num][1]=y;                memset(arc,false,sizeof(arc));                for (x=1;x<=n;x++)                   for (y=1;y<=m;y++)                      if ((x+y)%2 && s[x][y])                      {                             if (x!=1 && s[x-1][y]) arc[hash[x][y]][hash[x-1][y]]=true;                             if (y!=1 && s[x][y-1]) arc[hash[x][y]][hash[x][y-1]]=true;                             if (x!=n && s[x+1][y]) arc[hash[x][y]][hash[x+1][y]]=true;                             if (y!=m && s[x][y+1]) arc[hash[x][y]][hash[x][y+1]]=true;                      }                n=num;                if (cases) printf("\n");                cases++;                printf("%d\n",getmax());                for (i=1;i<=n;i++)                  if (match[i])                      printf("(%d,%d)--(%d,%d)\n",w[match[i]][0],w[match[i]][1],w[i][0],w[i][1]);       }       return 0;}