hdu4185

题解:
每两个联通的油井建边

然后二分图最大匹配

最后答案除以2

代码:

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=20005;
char c;
int T,a[N/20][N/20],f[N],match[N],fi[N],cas,num,ne[N],zz[N],n,cnt;
int dfs(int x)
{
    for (int i=fi[x];i;i=ne[i])
     if (!f[zz[i]])
      {
          f[zz[i]]=1;
          if (!match[zz[i]]||dfs(match[zz[i]]))
           {
               match[zz[i]]=x;
               return 1;
           }
      }
    return 0; 
}
void jb(int x,int y)
{
    ne[++num]=fi[x];
    fi[x]=num;
    zz[num]=y;
    ne[++num]=fi[y];
    fi[y]=num;
    zz[num]=x;
}
int main()
{
    scanf("%d",&T);
    while (T--)
     {
         memset(a,0,sizeof a);
         memset(fi,0,sizeof fi);
         num=0;
         scanf("%d",&n);
         for (int i=0;i<n;i++)
          for (int j=0;j<n;j++)
           {
               int ch=getchar();
               while (ch!='.'&&ch!='#')ch=getchar();
               if (ch=='#')a[i][j]=++cnt;
               else a[i][j]=0;
           }
         for (int i=0;i<n;i++)
         for (int j=0;j<n;j++)
          if (a[i][j])
           {
               if (a[i+1][j])jb(a[i][j],a[i+1][j]);
               if (a[i][j+1])jb(a[i][j],a[i][j+1]);
           } 
        int ans=0;   
        for (int i=1;i<=cnt;i++)
         {
             memset(f,0,sizeof f);
             ans+=dfs(i);
         }   
        printf("Case %d: %d
",++cas,ans/2); 
     } 
    return 0; 
}
原文地址:https://www.cnblogs.com/xuanyiming/p/8230025.html