Give you a matrix(only contains 0 or 1),every time you can select a row or a column and delete all the ‘1‘ in this row or this column . 

Your task is to give out the minimum times of deleting all the ‘1‘ in the matrix.

3 3 
0 0 0
1 0 1
0 1 0
0

2

题意：输入n行m列的数，这些数只有1或0组成，每次可以消掉一行或者一列的1，问，消灭掉全部的1最少需要进行多少次操作
思路：我二分图匹配模型还是没有转换过来，但是队友给我讲了以后，豁然开朗，矩阵的行x就相当于左集合，矩阵的列y就相当于
右集合，要使1被消灭，就相当于连通(x,y)，求只要找到最少的顶点把所有的边都覆盖了就是最少的操作次数。
sum没有初始化样例也能过，但是wrong了三次后才发现是这个原因，尴尬~~

#include<stdio.h>
#include<string.h>
#define N 110

int book[N],e[N][N],match[N];
int n,m;

int dfs(int u)
{
    int i;
    for(i = 1; i <= m; i ++)
    {
        if(!book[i]&&e[u][i])
        {
            book[i] = 1;
            if(!match[i]||dfs(match[i]))
            {
                match[i] = u;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int i,j,sum;
    while(scanf("%d",&n),n!=0)
    {
        scanf("%d",&m);
        memset(match,0,sizeof(match));
        memset(e,-1,sizeof(e));
        for(i = 1; i <= n; i ++)
            for(j = 1; j <= m; j ++)
                scanf("%d",&e[i][j]);
        sum = 0;
        for(i = 1; i <= n; i ++)
        {
            memset(book,0,sizeof(book));
            if(dfs(i))
                sum ++;
        }
        printf("%d\n",sum);
    }
    return 0;
}