HDU 1198 Farm Irrigation

题目大意：给你地图，让你判断需要多少水才可以将农场灌满。

题解：显然用并查集比较容易，将可以连通的并起来，最后输出连通块的数目即可，一开始我用字母分类讨论发现很麻烦，于是参考别人的博客发现，直接自己写一个矩阵，然后处理一下读入数据会比较简单：

#include <cstring>   

#include <cstdio>   

#include <iostream>  

using namespace std;  

int R[11][11]={{0,0,0,0,0,0,0,0,0,0,0},  

{1,0,1,0,0,1,1,1,1,0,1},{0,0,0,0,0,0,0,0,0,0,0},  

{1,0,1,0,0,1,1,1,1,0,1},{0,0,0,0,0,0,0,0,0,0,0},  

{1,0,1,0,0,1,1,1,1,0,1},{1,0,1,0,0,1,1,1,1,0,1},  

{0,0,0,0,0,0,0,0,0,0,0},{1,0,1,0,0,1,1,1,1,0,1},  

{1,0,1,0,0,1,1,1,1,0,1},{1,0,1,0,0,1,1,1,1,0,1}};  

int U[11][11]={{0,0,0,0,0,0,0,0,0,0,0},  

{0,0,0,0,0,0,0,0,0,0,0},{1,1,0,0,1,0,1,1,0,1,1},  

{1,1,0,0,1,0,1,1,0,1,1},{1,1,0,0,1,0,1,1,0,1,1},  

{0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0},  

{1,1,0,0,1,0,1,1,0,1,1},{1,1,0,0,1,0,1,1,0,1,1},  

{1,1,0,0,1,0,1,1,0,1,1},{1,1,0,0,1,0,1,1,0,1,1}};  

string map[55]; 

int f[3000];  

void init(int n)  

{  

    int i;  

    for(i=0;i<=n;i++)  

        f[i]=i;  

}  



int sf(int i)  

{  

    int j=i;  

    while(j!=f[j])  

    {  

        j=f[j];  

    }  

    return f[i]=j;  

}  



int Union(int x,int y)  

{  

    x=sf(x);  

    y=sf(y);  

    if(x==y)  

        return 0;  

    else  

    {  

        f[x]=y;  

        return 1;  

    }  

}  

 

int main()  

{  

    int n,m;  

    while(scanf("%d%d",&m,&n),m!=-1&&n!=-1)  

    {  

        int i,j;  

        init(n*m);  

        for(i=0;i<m;i++)  

            cin>>map[i];  

        for(i=0;i<m;i++)  

            for(j=1;j<n;j++)  

                if(R[map[i][j-1]-'A'][map[i][j]-'A'])  

                    Union(i*n+j-1,i*n+j);  

        for(i=0;i<n;i++)  

            for(j=1;j<m;j++)  

                if(U[map[j-1][i]-'A'][map[j][i]-'A'])  

                    Union((j-1)*n+i,j*n+i);  

        int count=0;  

        for(i=0;i<n*m;i++)  

        {  

            if(f[i]==i)  

                count++;  

        }  

        printf("%d\n",count);  

    }  

    return 0;  

}