Dinic

#include <bits/stdc++.h>
using namespace std;

const int maxn = 10010;
struct edge { int to, cap, rev; };  // 存储边的结构体 (终点, 容量, 反向边)
vector<edge> G[maxn];               // 邻接表
int level[maxn], iter[maxn];        // level 为到源点的距离编号, iter 为当前弧

// 添加一条从 from 到 to 的容量为 cap 的边
void add_edge(int from, int to, int cap) {
    G[from].push_back((edge){to, cap, G[to].size()});
    G[to].push_back((edge){from, 0, G[from].size() - 1});
}

// 计算从源点出发的距离
void bfs(int s) {
    memset(level, -1, sizeof(level));
    queue<int> q;
    level[s] = 0;
    q.push(s);
    while (!q.empty()) {
        int v = q.front();
        q.pop();
        for (int i = 0; i < G[v].size(); i++) {
            edge &e = G[v][i];
            if (e.cap > 0 && level[e.to] == -1) {
                level[e.to] = level[v] + 1;
                q.push(e.to);
            }
        }
    }
}

// dfs 寻找增广路
int dfs(int v, int t, int f) {
    if (v == t) {
        return f;
    }
    for (int &i = iter[v]; i < G[v].size(); i++) {
        edge &e = G[v][i];
        if (e.cap > 0 && level[v] < level[e.to]) {
            int d = dfs(e.to, t, min(f, e.cap));
            if (d > 0) {
                e.cap -= d;
                G[e.to][e.rev].cap += d;
                return d;
            }
        }
    }
    return 0;
}

// 求从 s 到 t 的最大流
int dinic(int s, int t) {
    int flow = 0;
    while (true) {
        bfs(s);
        if (level[t] < 0) {
            return flow;
        }
        memset(iter, 0, sizeof(iter));
        int f;
        while ((f = dfs(s, t, 0x3f3f3f3f)) > 0) {
            flow += f;
        }
    }
}

int main() {
    ios::sync_with_stdio(false);
    int n, m, s, t;
    cin >> n >> m >> s >> t;
    while (m--) {
        int from, to, cap;
        cin >> from >> to >> cap;
        add_edge(from, to, cap);
    }
    cout << dinic(s, t) << endl;
    return 0;
}