HDU 2119 Matrix

Matrix

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3057    Accepted Submission(s): 1402

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

 二分匹配水题

#include <iostream> 
#include <algorithm> 
#include <cstring> 
#include <cstdio>
#include <vector> 
#include <queue> 
#include <cstdlib> 
#include <iomanip>
#include <cmath> 
#include <ctime> 
#include <map> 
#include <set> 
using namespace std; 
#define lowbit(x) (x&(-x)) 
#define max(x,y) (x>y?x:y) 
#define min(x,y) (x<y?x:y) 
#define MAX 100000000000000000 
#define MOD 1000000007
#define pi acos(-1.0) 
#define ei exp(1) 
#define PI 3.141592653589793238462
#define ios() ios::sync_with_stdio(false)
#define INF 0x3f3f3f3f 
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
int a[110][110];
int n,m,x,y;
int vis[109];
int pos[109];
bool find(int x)
{
    for(int i=1;i<=m;i++)
    {
        if(a[x][i] && !vis[i])
        {
            vis[i]=1;
            if(!pos[i] || find(pos[i]))
            {
                pos[i]=x;
                return true;
            }
        }
    }
    return false;
}
int main()
{
    int cast=0;
    while(scanf("%d",&n)&&n)
    {
        scanf("%d",&m);
        int ans=0;
        memset(pos,0,sizeof(pos));
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
               scanf("%d",&a[i][j]);
            }
        }
        for(int i=1;i<=n;i++)
        {
            memset(vis,0,sizeof(vis));
            if(find(i)) ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}