HDU-1845-Jimmy's Assignment

链接：https://vjudge.net/problem/HDU-1845

题意：

给一个有向图，求最大匹配。

思路：

有相图的最大匹配，可以通过加上反向边， 求这个无向图的最大匹配，

原图的最大匹配就是无向图的最大匹配除2.

详细解释：https://xwk.iteye.com/blog/2129301

https://blog.csdn.net/u013480600/article/details/38638219

代码：

#include <iostream>
#include <memory.h>
#include <string>
#include <istream>
#include <sstream>
#include <vector>
#include <stack>
#include <algorithm>
#include <map>
#include <queue>
#include <math.h>
#include <cstdio>
#include <set>
#include <iterator>
#include <cstring>
using namespace std;

typedef long long LL;
const int MAXN = 5e3+10;
int Next[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};

vector<int> G[MAXN];
int Link[MAXN], Vis[MAXN];
int n, m, k;

bool Dfs(int x)
{
    for (int i = 0;i < G[x].size();i++)
    {
        int node = G[x][i];
        if (Vis[node] == 0)
        {
            Vis[node] = 1;
            if (Link[node] == -1 || Dfs(Link[node]))
            {
                Link[node] = x;
                return true;
            }
        }
    }
    return false;
}

int Solve()
{
    memset(Link, -1, sizeof(Link));
    int cnt = 0;
    for (int i = 1;i <= n;i++)
    {
        memset(Vis, 0, sizeof(Vis));
        if (Dfs(i))
            cnt++;
    }
    return cnt;
}

int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d", &n);
        for (int i = 1;i <= n;i++)
            G[i].clear();
        int l, r;
        for (int i = 1;i <= 3*n/2;i++)
        {
            scanf("%d%d", &l, &r);
            G[l].push_back(r);
            G[r].push_back(l);
        }
        int cnt = Solve();
        printf("%d
", cnt/2);
    }

    return 0;
}