Matrix

Problem Description

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

题意描述：

将一个只包含0或1的矩阵删除所有的1，每次可以选择一行或一列，能够删除这一行或一列的1，求删除所有1的最小次数。

解题思路：

二分匹配模板题求最大的匹配数，利用二分匹配求出最大匹配数即为删除所有1的最小次数。

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