#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for (int i = a; i < (b); i++)
#define Rep(i, a, b) for (int i = a; i >= (b); i--)
#define pb push_back
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define fi first
#define se second
#define SZ(x) (int)(x.size())
#define lowbit(x) ((x) & -(x))
using db = double;
using ll = long long;
using ull = unsigned long long;
using vi = vector<int>;
using vvi = vector<vi>;
using pii = pair<int, int>;
using i128 = __int128_t;
// Think twice, code once

namespace Dinic {
const int N = ;
const int inf = ;
struct Edge {
    int from, to, cap, flow;
};

bool operator<(const Edge &a, const Edge &b) {
    return a.from < b.from || (a.from == b.from && a.to < b.to);
}

int n, m, s, t;
vector<Edge> edges;  // 边数的两倍
vector<int> G[N];    // 邻接表，G[i][j]表示结点i的第j条边在e数组中的序号
bool vis[N];         // BFS使用
int d[N];            // 从起点到i的距离
int cur[N];          // 当前弧指针

void init(int _n) {
    n = n;
    for (int i = 0; i < n; i++) G[i].clear();
    edges.clear();
}

void add(int from, int to, int cap) {
    edges.push_back((Edge){from, to, cap, 0});
    edges.push_back((Edge){to, from, 0, 0});
    m = edges.size();
    G[from].push_back(m - 2);
    G[to].push_back(m - 1);
}

bool BFS() {  //  最后一次bfs走过的都是交错路径,可用于二分图博弈判断H点是否必须
    memset(vis, 0, sizeof(int) * n);
    queue<int> Q;
    Q.push(s);
    vis[s] = 1;
    d[s] = 0;
    while (!Q.empty()) {
        int x = Q.front();
        Q.pop();
        for (int i = 0; i < G[x].size(); i++) {
            Edge &e = edges[G[x][i]];
            if (!vis[e.to] && e.cap > e.flow) {
                vis[e.to] = 1;
                d[e.to] = d[x] + 1;
                Q.push(e.to);
            }
        }
    }
    return vis[t];
}

int DFS(int x, int a) {
    if (x == t || a == 0) return a;
    int flow = 0, f;
    for (int &i = cur[x]; i < G[x].size(); i++) {
        Edge &e = edges[G[x][i]];
        if (d[x] + 1 == d[e.to] &&
            (f = DFS(e.to, min(a, e.cap - e.flow))) > 0) {
            e.flow += f;
            edges[G[x][i] ^ 1].flow -= f;
            flow += f;
            a -= f;
            if (a == 0) break;
        }
    }
    return flow;
}

int Maxflow(int _s, int _t) {
    s = _s, t = _t;
    int flow = 0;
    while (BFS()) {
        memset(cur, 0, sizeof(int) * n);
        flow += DFS(s, inf);
    }
    return flow;
}
}  // namespace Dinic
void solve() {
    int n, m, s, t; cin >> n >> m >> s >> t; s--, t--;
    Dinic::init(n);
    while (m--) {
        int u, v, c; cin >> u >> v >> c; u--, v--;
        Dinic::add(u, v, c);
    }
    cout << Dinic::Maxflow(s, t) << endl;
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    int T = 1;
    // cin >> T;
    while (T--) solve();
    return 0;
}

answer.code:22:15: error: expected primary-expression before ‘;’ token
   22 | const int N = ;
      |               ^
answer.code:23:17: error: expected primary-expression before ‘;’ token
   23 | const int inf = ;
      |                 ^
answer.code:34:15: error: size of array ‘G’ is not an integral constant-expression
   34 | vector<int> G[N];    // 邻接表，G[i][j]表示结点i的第j条边在e数组中的序号
      |               ^
answer.code:35:10: error: size of array ‘vis’ is not an integral constant-expression
   35 | bool vis[N];         // BFS使用
      |          ^
answer.code:36:7: error: size of array ‘d’ is not an integral constant-expression
   36 | int d[N];            // 从起点到i的距离
      |       ^
answer.code:37:9: error: size of array ‘cur’ is not an integral constant-expression
   37 | int cur[N];          // 当前弧指针
      |         ^