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

题意：

    就是每次可以删除一行或者一列数，问最少几次可以把所有的1都变成0，也就是都删完。

思路：

       把行号放在二分图左边点集,列号放在右边点集.如果(i,j)格子是1,那么就连左i与右j的无向边。

       就是求该二分图的最小边覆蓋点集.使得每条边都被至少1个点覆蓋。

       最小覆蓋集大小 = 最大匹配数。

代码：

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