hdoj 2119 Matrix 【行列匹配 求解最小点覆盖】【基础题】

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 3 3 0 0 0 1 0 1 0 1 0 0   Sample Output 2

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;
}