InputThere are several test cases.

The first line contains two integers n,m(1<=n,m<=100), n is the number of rows of the given matrix and m is the number of columns of the given matrix.
The next n lines describe the matrix:each line contains m integer, which may be either ‘1’ or ‘0’.

n=0 indicate the end of input.
OutputFor each of the test cases, in the order given in the input, print one line containing the minimum times of deleting all the ‘1‘ in the matrix.
Sample Input

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

Sample Output

2

題意：輸入n行m列的數，這些數只有1或0組成，每次可以消掉一行或者一列的1，問，消滅掉全部的1最少需要進行多少次操作
思路：我二分圖匹配模型還是沒有轉換過來，但是隊友給我講了以後，豁然開朗，矩陣的行x就相當於左集合，矩陣的列y就相當於
右集合，要使1被消滅，就相當於連通(x,y)，求只要找到最少的頂點把所有的邊都覆蓋了就是最少的操作次數。
sum沒有初始化樣例也能過，但是wrong了三次後才發現是這個原因，尷尬~~

#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;
}

【二分圖匹配入門專題1】D - Matrix hdu2119【最小頂點覆蓋】