hdu 2119 Matrix-二分图

poj 3041

F - Matrix

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit   Status   Practice   HDU 2119

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
        
        
        
        

 

就是求覆盖所以1的最少的行加列的数量呢~！

二分法做。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string>
#include<cstring>
#include<cmath>
#include<queue>
using namespace std;
#define sf scanf
#define pf printf
#define INF 1<<29
#define clr(x) memset(x,0,sizeof(x))
#define Clr(x) memset(x,-1,sizeof(x));
#define N 500
int map[N][N];
int vis[N];
int link[N];
int n,m;
int find(int x){
	for(int i=1;i<=m;i++){
		if(map[x][i]&&!vis[i]){
			vis[i]=1;
			if(!link[i]||find(link[i])){
				link[i]=x;
				return true;
			}
		}
	}
	return false;
}
int main(){
	int sum;
	while(~sf("%d",&n)&&n){
		sf("%d",&m);
		clr(link);
		for(int i=1;i<=n;i++)
			for(int j=1;j<=m;j++)
				sf("%d",&map[i][j]);
		sum=0;
		for(int i=1;i<=n;i++){
			clr(vis);
			sum+=find(i);
		}
		pf("%d\n",sum);
	}
	return 0;
}