zoj1654

题解：

对于每一联通的x，y

检点

然后交叉的连边

然后二分图

代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int N=;
int cas,n,m,num,t,x[N][N],y[N][N],f[N],fi[N],ne[N],zz[N],match[N],r,c;
char s[N][N];
void jb(int x,int y)
{
    zz[++num]=y;
    ne[num]=fi[x];
    fi[x]=num;
}
void build()
{
    r=,c=;
    for (int i=;i<n;i++)
     for (int j=;j<m;j++)
      {
        if (s[i][j]!='#'&&!x[i][j])
         {
            for (int k=j;s[i][k]!='#'&&k<m;k++){x[i][k]=r;}
            r++;
         }
        if (s[i][j]!='#'&&!y[i][j])
         {
            for (int k=i;s[k][j]!='#'&&k<n;k++){y[k][j]=c;}
            c++;
         }
      }
    for (int i=;i<n;i++)
     for (int j=;j<m;j++)
      if (s[i][j]=='o'&&x[i][j]&&y[i][j])jb(x[i][j],y[i][j]);
}
int dfs(int x)
{
    for (int i=fi[x];i;i=ne[i])
     if (!f[zz[i]])
      {
          f[zz[i]]=;
          if (!match[zz[i]]||dfs(match[zz[i]]))
           {
               match[zz[i]]=x;
               return ;
           }
      }
    return ;
}
int main()
{
    scanf("%d",&t);
    while (t--)
     {
            memset(x,,sizeof x);
        memset(y,,sizeof y);
        memset(fi,,sizeof fi);
        memset(match,,sizeof match);
        num=;
        scanf("%d%d",&n,&m);
        for (int i=;i<n;i++)scanf("%s",s[i]);
            build();
            printf("Case :%d\n",++cas);
        int ans=;
        for (int i=;i<=r;i++)
         {
            memset(f,,sizeof f);
               ans+=dfs(i);
          }
        printf("%d\n",ans);
     }
}