二分图匹配

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <deque>
#include <bitset>
#include <cassert>
//#include <unordered_set>
//#include <unordered_map>
#define ll              long long
#define pii             pair<int, int>
#define rep(i,a,b)      for(int  i=a;i<=b;i++)
#define dec(i,a,b)      for(int  i=a;i>=b;i--)
#define forn(i, n)      for(int i = 0; i < int(n); i++)
using namespace std;
int dir[4][2] = { { 1,0 },{ 0,1 } ,{ 0,-1 },{ -1,0 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-6;
const int mod = 1e9 + 7;

inline ll read()
{
    ll x = 0; bool f = true; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? x : -x;
}
inline ll gcd(ll m, ll n)
{
    return n == 0 ? m : gcd(n, m % n);
}
void exgcd(ll A, ll B, ll& x, ll& y)
{
    if (B) exgcd(B, A % B, y, x), y -= A / B * x; else x = 1, y = 0;
}
inline int qpow(int x, ll n) {
    int r = 1;
    while (n > 0) {
        if (n & 1) r = 1ll * r * x % mod;
        n >>= 1; x = 1ll * x * x % mod;
    }
    return r;
}
inline int inv(int x) {
    return qpow(x, mod - 2);
}
ll lcm(ll a, ll b)
{
    return a * b / gcd(a, b);
}
/**********************************************************/
const int N = 2e5 + 5;
struct augment_path {
    vector<vector<int> > g;
    vector<int> pa;  // 匹配
    vector<int> pb;
    vector<int> vis;  // 访问
    int n, m;         // 顶点和边的数量
    int dfn;          // 时间戳记
    int res;          // 匹配数

    augment_path(int _n, int _m) : n(_n+1), m(_m+1) {
        assert(0 <= n && 0 <= m);
        pa = vector<int>(n, -1);
        pb = vector<int>(n, -1);
        vis = vector<int>(n);
        g.resize(n);
        res = 0;
        dfn = 0;
    }

    void add(int from, int to) {
        //assert(0 <= from && from < n && 0 <= to && to < m);
        g[from].push_back(to);
    }

    bool dfs(int v) {
        vis[v] = dfn;
        for (int u : g[v]) {
            if (pb[u] == -1) {
                pb[u] = v;
                pa[v] = u;
                return true;
            }
        }
        for (int u : g[v]) {
            if (vis[pb[u]] != dfn && dfs(pb[u])) {
                pa[v] = u;
                pb[u] = v;
                return true;
            }
        }
        return false;
    }

    int solve() {
        while (true) {
            dfn++;
            int cnt = 0;
            for (int i = 1; i < n; i++) {
                if (pa[i] == -1 && dfs(i)) {
                    cnt++;
                }
            }
            if (cnt == 0) {
                break;
            }
            res += cnt;
        }
        return res;
    }
};

int main()
{
    int nl, nr, m;
    cin >> nl >> nr >> m;
    augment_path s(nl+nr,m);
    rep(i, 1, m)
    {
        int u, v;
        cin >> u >> v;
        v += nl;
        s.add(u, v);
        s.add(v, u);
    }
    int res = s.solve();
    cout << res/2 << endl;
    return 0;
}
原文地址:https://www.cnblogs.com/dealer/p/13585482.html