算法笔记--匈牙利算法

http://www.renfei.org/blog/bipartite-matching.html

匈牙利算法:从左边的点出发不断地搜索,寻找增广路,交换交替路上的路径,来求解二分图的最大匹配。

模板:

bool dfs(int u) {
    for (int v : g[u]) {
        if(!vis[v]) {
            vis[v] = true;
            if(match[v] == -1 || dfs(match[v])) {
                match[v] = u;
                match[u] = v;
                return true;
            }
        }
    }
    return false;
}
int ans = 0;
mem(match, -1);
for (int i = 1; i <= n; ++i) {
    if(match[i] == -1) {
        mem(vis, false);
        if(dfs(i)) ans++;
  }
}

P3386 【模板】二分图匹配 

代码:

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define y1 y11
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define pdd pair<double, double>
#define mem(a, b) memset(a, b, sizeof(a))
#define debug(x) cerr << #x << " = " << x << "\n";
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//head

const int N = 2e3 + 5;
int match[N];
bool vis[N];
vector<int> g[N];
int n, m, e, u, v;
bool dfs(int u) {
    for (int v : g[u]) {
        if(!vis[v]) {
            vis[v] = true;
            if(match[v] == -1 || dfs(match[v])) {
                match[v] = u;
                match[u] = v;
                return true;
            }
        }
    }
    return false;
}
int main() {
    scanf("%d %d %d", &n, &m, &e);
    for (int i = 1; i <= e; ++i) {
        scanf("%d %d", &u, &v);
        if(u > n || v > m) continue;
        g[u].pb(v+n);
        g[v+n].pb(u);
    }
    int ans = 0;
    mem(match, -1);
    for (int i = 1; i <= n; ++i) {
        if(match[i] == -1) {
            mem(vis, false);
            if(dfs(i)) ans++;
        }
    }
    printf("%d\n", ans);
    return 0;
}
View Code

 

posted @ 2019-03-28 17:36  Wisdom+.+  阅读(189)  评论(0编辑  收藏  举报