acwing 861 Maximum matching of bipartite graphs (Hungarian algorithm)

Topic

answer

1. Hungarian algorithm:Find the maximum number of matches of a bipartite graph, O(nm), which is actually much smaller than this time complexity, and follows the principle of "first come first serve, let you let"

2. Simulation process: 1 matches the first 6, 2 matches the first 7, and 3 has only one 6 that can match, but 6 has matched 1, we need to see whether the point 1 matched by 6 can match other points Point, found that 1 can match 8, in addition to matching 6, so let 1 match 8, 3 match 6, 4 match 7.

Code

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>

using namespace std;
const int N = 1e5 + 10;


int n1, n2, m;
int h[N], e[N], ne[N], idx;
int match[N];  //存储每个点当前匹配的点
bool st[N];    //表示每个点是否已经被遍历过

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

bool find(int x) {
    
    

    for (int i = h[x]; i != -1; i=ne[i]) {
    
    
        int j = e[i];
        if (!st[j]) {
    
    
            st[j] = true;
            //如果这个点没有被匹配或者是这个点所匹配的点能找到别的点
            if(match[j]==0||find(match[j])){
    
    
                match[j]=x;
                return true;
            }
        }
    }
    return false;
}

int main() {
    
    

    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);

    memset(h, -1, sizeof h);

    cin >> n1 >> n2 >> m;

    for (int i = 1; i <= m; i++) {
    
    
        int a, b;
        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;
}