【匈牙利算法】AcWing861.二分图的最大匹配

AcWing861.二分图的最大匹配

题解

匈牙利算法：若该左点匹配到的右点之前已匹配到了左点，则查找该之前匹配的左点可否换一个点，使新的左点能匹配该右点

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

using namespace std;

const int N = 510, M = 1e5 + 10;

bool st[N];
int match[N];
int n1, n2, m, h[N], ne[M], e[M], idx;

void add(int a, int b)
{
    ne[idx] = h[a], h[a] = idx, e[idx++] = b; 
}

bool find(int u)
{
    for(int i = h[u]; ~i; i = ne[i])
    {
        int j = e[i];
        if(!st[j])
        {
            st[j] = true;
            if(!match[j] || find(match[j]))
            {
                match[j] = u;
                return true;
            }
        }
    }
    return false;
}

int main()
{
    memset(h, -1, sizeof h);
    cin >> n1 >> n2 >> m;
    int a, b;
    for(int i = 0; i < m; ++i)
    {
        cin >> a >> b;
        add(a, b);
    }
    int res = 0;
    for(int i = 1; i <= n1; ++i)
    {
        memset(st, false, sizeof st);
        if(find(i))
            res++;
    }
    cout << res << endl;
    return 0;
}