Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1076 Accepted Submission(s): 448
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
//行列(一整行，一整列)拆分成 X,Y集合，编号1-n 和 1-m
//某一行列相交有1时，X集合行编号和Y集合相应编号建立连接关系
//这时二分图的边就对应原图中的1 转化成求最小点覆盖、、、就是求最大匹配
#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
#define N 102
int map[N][N],visit[N],match[N];
int n,m;
bool DFS(int k)
{
for(int i=1;i<=m;i++)
{
if(visit[i]||!map[k][i])
continue;
visit[i]=1;
if(!match[i]||DFS(match[i]))
{
match[i]=k;
return true;
}
}
return false;
}
int main()
{
int i,j;
while(scanf("%d",&n),n)
{
scanf("%d",&m);
for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
scanf("%d",&map[i][j]);
memset(match,0,sizeof(match));
int cnt=0;
for(i=1;i<=n;i++)
{
memset(visit,0,sizeof(visit));
if(DFS(i)) cnt++;
}
printf("%d\n",cnt);
}
return 0;
}