卡常：

#include<stdio.h>
#define N 209
#define M 5009

struct node
{
	int v, w, nxt;
}e[M << 1];

int head[N], cnt = 1;

inline void add(int u, int v, int w)
{
	e[++ cnt].v = v, e[cnt].w = w, e[cnt].nxt = head[u], head[u] = cnt;
}

inline int read()
{
	register int ret = 0;
	register bool f = 1;
	register char ch = getchar();
	while (ch < '0' || ch > '9')
		(ch == '-') ? f = 0 : 0, ch = getchar();
	while (ch >= '0' && ch <= '9')
		ret = (ret << 1) + (ret << 3) + (ch ^ 48), ch = getchar();
	return f ? ret : -ret;
}

int s, t;

int dis[N], now[N];

#include<cstring>

using std::memset;

#include<queue>

using std::queue;

inline bool bfs()
{
	memset(dis, -1, sizeof dis);
	register int u, v, w, i;
	queue < int > q;
	q.push(s);
	dis[s] = 0;
	now[s] = head[s];
	while (!q.empty())
	{
		u = q.front();
		q.pop();
		for (i = head[u];i;i = e[i].nxt)
		{
			v = e[i].v, w = e[i].w;
			if (w > 0 && dis[v] == -1)
			{		
				dis[v] = dis[u] + 1;
				now[v] = head[v];
				if (v == t)
					return 1;
				q.push(v);
			}
		}
	}
	return 0;
}

inline int min(int a, int b)
{
	return a < b ? a : b;
}

#define ll long long

int dfs(register int u, register int flow = 1e9)
{
	if (u == t)
		return flow;
	register int lost = flow, v, w, k;
	for (register int i = now[u];i && lost;i = e[i].nxt, now[u] = i)
	{
		v = e[i].v, w = e[i].w;
		if (dis[v] == dis[u] + 1 && w > 0)
		{
			k = dfs(v, min(lost, w));
			if (!k)
			{
				dis[v] = -1;
				continue;
			}
			lost -= k;
			e[i].w -= k;
			e[i ^ 1].w += k;
		}
	}
	return flow - lost;
}

int main()
{
	register int n = read(), m = read(), u, v, w;
	register ll ans = 0;
	s = read(), t = read();
	while (m --)
		u = read(), v = read(), w = read(), add(u, v, w), add(v, u, 0);
	while (bfs())
		ans += dfs(s);
	printf("%lld", ans);
	
	return 0;
}

未卡常：

#include<stdio.h>
#define N 409
#include<queue>
using namespace std;
#define M 10009

int dis[N];

struct node
{
	int v, w, nxt;
}e[M << 1];

int head[N], cnt = 1, now[N];

inline void add(int u, int v, int w)
{
	e[++ cnt].v = v, e[cnt].w = w, e[cnt].nxt = head[u], head[u] = cnt;
}

inline int read()
{
	register int ret = 0, f = 1;
	register char ch = getchar();
	while (ch < '0' || ch > '9')
		(ch == '-') ? f = -1 : 0, ch = getchar();
	while (ch >= '0' && ch <= '9')
		ret = (ret << 1) + (ret << 3) + (ch ^ 48), ch = getchar();
	return ret * f;
}

int n, s, t;

inline bool bfs()
{
	for (register int i = 1;i <= n;++ i)
		dis[i] = -1;
	queue < int > q;
	dis[s] = 0;
	q.push(s);
	now[s] = head[s];
	while (!q.empty())
	{
		register int u = q.front();
		q.pop();
		for (register int i = head[u];i;i = e[i].nxt)
			if (e[i].w > 0 && dis[e[i].v] == -1)//构建分层图
			{
				int v = e[i].v;
				now[v] = head[v];
				dis[v] = dis[u] + 1;
				if (v == t)
					return 1;//找到有路径
				q.push(v);
			}
	}
	return 0;//没找到
}

inline long long min(long long a, long long b) {return a < b ? a : b;}

long long dfs(register int u, register long long sum = 1e9)
{
	if (u == t)
		return sum;
	register long long lost = sum;
	for (register int i = now[u];i && lost;i = e[i].nxt)
	{
		now[u] = i;
		int v = e[i].v, w = e[i].w;
		if (dis[v] == dis[u] + 1 && w > 0)//保证不是同一个层并且可以流动
		{
			int k = dfs(v, min(w, lost));//尽量让少的去流，也就是 EK 的想法
			if (!k)//没有，那就真没有
			{
				dis[e[i].v] = -1;
				continue;
			}
			lost -= k;
			e[i].w -= k;
			e[i ^ 1].w += k;
		}
	}
	return sum - lost;
}

inline long long Dinic()
{
	register long long ret = 0;
	while (bfs())
		ret += dfs(s);
	return ret;
}

int main()
{
	n = read();
	register int m = read();
	s = read(), t = read();
	register int u, v, w;
	while (m --)
		u = read(), v = read(), w = read(), add(u, v, w), add(v, u, 0);
	printf("%lld", Dinic());
	
	return 0;
}