费用流（模板）

//#include<bits/stdc++.h>
#include<algorithm>
#include <iostream>
#include   <fstream>
#include  <cstdlib>
#include  <cstring>
#include  <cassert>
#include   <cstdio>
#include   <vector>
#include   <string>
#include    <cmath>
#include    <queue>
#include    <stack>
#include      <set>
#include      <map>
using namespace std;
#define P(a,b,c) make_pair(a,make_pair(b,c))
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n;i>=a;i--)
#define CLR(vis) memset(vis,0,sizeof(vis))
#define MST(vis,pos) memset(vis,pos,sizeof(vis))
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef pair<int,pair<int,int> >pii;
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1000000007;
ll gcd(ll a, ll b) {
	return b ? gcd(b, a%b) : a;
}
template<class T>inline void gmax(T &A, T B) {
	(A<B)&&(A=B);//if(B>A)A=B;
}
template<class T>inline void gmin(T &A, T B) {
	(A>B)&&(A=B);//if(B<A)A=B;
}
template <class T>
inline bool scan_d(T &ret) {
	char c;
	int sgn;
	if(c=getchar(),c==EOF) return 0; //EOF
	while(c!='-'&&(c<'0'||c>'9')) c=getchar();
	sgn=(c=='-')?-1:1;
	ret=(c=='-')?0:(c-'0');
	while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
	ret*=sgn;
	return 1;
}
inline void outt(int x) {
	if(x>9) outt(x/10);
	putchar(x%10+'0');
}
const int MX = 505;
const int inf = 0x3f3f3f3f;
const int MXE = MX * MX * 4;
struct MinCost_MaxFlow {
	struct Edge {
		int v, w, nxt;
		int cost;
	} E[MXE];
	int head[MX], tot, level[MX], pre[MX], d[MX];
	bool vis[MX];
	void init() {
		memset(head, -1, sizeof(head));
		tot = 0;
	}
	void add(int u, int v, int w, int cost) {
		E[tot].v = v;
		E[tot].w = w;
		E[tot].cost = cost;
		E[tot].nxt = head[u];
		head[u] = tot++;
		E[tot].v = u;
		E[tot].w = 0;
		E[tot].cost = -cost;
		E[tot].nxt = head[v];
		head[v] = tot++;
	}
	bool spfa(int s, int t) {
		memset(vis, 0, sizeof(vis));
		memset(d, 0x3f, sizeof(d));
		memset(pre, -1, sizeof(pre));
		queue<int>q;
		q.push(s);
		d[s] = 0;
		vis[s] = 1;
		while (!q.empty()) {
			int u = q.front();
			q.pop();
			vis[u] = 0;
			for (int i = head[u]; ~i; i = E[i].nxt) {
				int w = E[i].w, v = E[i].v, cost = E[i].cost;
				if (w > 0 && d[v] > d[u] + cost) {
					d[v] = d[u] + cost;
					pre[v] = i;
					if (!vis[v]) {
						q.push(v);
						vis[v] = 1;
					}
				}
			}
		}
		//如果是最小费用可行流则要这一句(要求费用最小,不要求流量最大)
		//if (d[t] > 0) return false;
		return pre[t] != -1;
	}
	int solve(int s, int t, int &cost) {
		int flow = 0;
		cost = 0;
		while (spfa(s, t)) {
			int minFlow = inf;
			for (int i = pre[t]; ~i; i = pre[E[i ^ 1].v])
				minFlow = min(minFlow, E[i].w);
			for (int i = pre[t]; ~i; i = pre[E[i ^ 1].v]) {
				cost += minFlow * E[i].cost;
				E[i].w -= minFlow;
				E[i ^ 1].w += minFlow;
			}
			flow += minFlow;
		}
		return flow;
	}
} F;

int main() {
	F.init();
	s=?,t=?;
	add(u,v,flow,Cost);
	int cost = 0;
	F.solve(s, t, cost);
	printf("%d\n", cost);
	return 0;
}