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HDU2119_Matrix(二分图/最小点覆盖=最大匹配)

2024-07-17 20:30:17   219人阅读

解题报告

题目传送门

题意：

题意类似与POJ3041

思路：

见POJ3041解题报告

最小点覆盖。

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;
int mmap[110][110],vis[110],pre[110],n,m;
int dfs(int x) {
    for(int i=1; i<=m; i++) {
        if(!vis[i]&&mmap[x][i]) {
            vis[i]=1;
            if(pre[i]==-1||dfs(pre[i])) {
                pre[i]=x;
                return 1;
            }
        }
    }
    return 0;
}
int main() {
    int t,i,j;
    while(~scanf("%d",&n)) {
        if(!n)break;
        scanf("%d",&m);
        memset(pre,-1,sizeof(pre));
        memset(mmap,0,sizeof(mmap));
        for(i=1; i<=n; i++)
            for(j=1; j<=m; j++)
                scanf("%d",&mmap[i][j]);
        int ans=0;
        for(i=1; i<=n; i++) {
            memset(vis,0,sizeof(vis));
            ans+=dfs(i);
        }
        printf("%d\n",ans);
    }
    return 0;
}

Matrix

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1882    Accepted Submission(s): 833


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
 


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