Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2319    Accepted Submission(s): 1030

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

 

Author

Wendell

 

Source

给了一个n*m的矩阵，矩阵元素有0 1，每次可以消除一行或者一列的1，问最少多少次消除全部的1

最少点覆盖=最大匹配，匈牙利算法：

#include<stdio.h>

#include<vector>

#include<string.h>

#include<algorithm>

using namespace std;

vector<int>map[200];

int pipei[200],used[200];

int find(int x)

{

	for(int i=0;i<map[x].size();i++)

	{

		int y=map[x][i];

		if(!used[y])

		{

			used[y]=1;

			if(!pipei[y]||find(pipei[y]))

			{

				pipei[y]=x;

				return 1;

			}

		}

	}

	return 0;

}

int main()

{

	int n,m;

	while(scanf("%d",&n),n)

	{

		scanf("%d",&m);

		for(int i=1;i<=n;i++)

		{

			map[i].clear();

			pipei[i]=0;

		}

		int a;

		for(int i=1;i<=n;i++)

		{

			for(int j=1;j<=m;j++)

			{

				scanf("%d",&a);

				if(a)

					map[i].push_back(j);

			}

		}

		int sum=0;

		for(int i=1;i<=n;i++)

		{

			memset(used,0,sizeof(used));

			sum+=find(i);

		}

		printf("%d\n",sum);

	}

}