HDU-2119-Matrix（二分匹配）

                                                             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：

给你一个0,1矩阵，让你每次可以将一行或者一列的1变成0，让你求最少几次可以让所有的1都变成0，将矩阵中是1的格子的x和y坐标记录下来，然后再求最大匹配数

程序代码：

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