hdoj 2119 Matrix 【行列匹配求解最小点覆盖】【基础题】

最新推荐文章于 2019-05-08 21:51:39 发布

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最大匹配(权值匹配)、独立集、团 && 最小路径、点覆盖专栏收录该内容

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本文介绍了一道关于矩阵操作的问题，通过使用匈牙利算法来解决如何以最少的操作次数消除矩阵中所有1的问题。文章提供了完整的代码实现。

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Matrix

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2234 Accepted Submission(s): 992

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

Sample Output

题意：给你一个N*M的矩阵，矩阵里面有一些位置是1。每次操作可以划去同一行或者同一列的1，问你最少需要几次操作。

水题吧，最小点覆盖 = 最大匹配。直接一次匈牙利就ok了。

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAXN 101
using namespace std;
int Map[MAXN][MAXN];
int pipei[MAXN];
bool used[MAXN];
int N, M;
void getMap()
{
    int a;
    memset(Map, 0, sizeof(Map));
    for(int i = 1; i <= N; i++)
    {
        for(int j = 1; j <= M; j++)
        {
            scanf("%d", &a);
            if(a)
                Map[i][j] = 1;//建立关系
        }
    }
}
int DFS(int x)
{
    for(int i = 1; i <= M; i++)
    {
        if(Map[x][i] && !used[i])
        {
            used[i] = true;
            if(pipei[i] == -1 || DFS(pipei[i]))
            {
                pipei[i] = x;
                return 1;
            }
        }
    }
    return 0;
}
void solve()
{
    int ans = 0;
    memset(pipei, -1, sizeof(pipei));
    for(int i = 1; i <= N; i++)
    {
        memset(used, false, sizeof(used));
        ans += DFS(i);
    }
    printf("%d\n", ans);
}
int main()
{
    while(scanf("%d", &N), N)
    {
        scanf("%d", &M);
        getMap();
        solve();
    }
    return 0;
}