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.
Input
    There 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.
Output
    For 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：給你一個0,1矩陣，讓你每次可以將一行或者一列的把1變成0，求最少幾次可以讓所有的1都變成0。

分析：二分匹配題，將矩陣中是1的格子的x和y座標記錄下來，然後再求最大匹配數即可。

#include<stdio.h>
#include<string.h>
int m,n,e[110][110],match[110],book[110];
int dfs(int u)
{
	int i;
	for(i = 1; i <= m; i ++)
	{
		if(book[i] == 0 && e[u][i] == 1)
		{
			book[i] = 1;
			if(match[i] == 0 || dfs(match[i]))
			{
				match[i] = u;
				return 1;
			}
		}
	}
	return 0;
}
int main()
{
	int i,j,k,sum;
	while(scanf("%d",&n), n != 0)
	{
		scanf("%d",&m);
		sum = 0;
		memset(e,0,sizeof(e));
		memset(match,0,sizeof(match));
		for(i = 1; i <= n; i ++)
			for(j = 1; j <= m; j ++)
			{
				scanf("%d",&k);
				if(k == 1)
					e[i][j] = 1;
			}
		for(i = 1; i <= n; i ++)
		{
			memset(book,0,sizeof(book));
			if(dfs(i))
				sum ++;
		}
		printf("%d\n",sum);	
	}
	return 0;
}