Aimee

最小费用是最大流基础上的

那么就用spfa代替Ek的bfs就行

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
int n,m,s,t;
int u,v,w,c;
const int maxn=5001;
const int maxm=50001;
int head[maxn];
struct b{
	int to;
	int ne;
	int v;
	int c;
}ed[maxm*2];
 int p=1;
 void add(int f,int t,int v,int c){
	ed[++p].ne=head[f];ed[p].to=t;ed[p].v=v;head[f]=p;ed[p].c=c;
	ed[++p].ne=head[t];ed[p].to=f;ed[p].v=0;head[t]=p;ed[p].c=-c;
}
queue<int> q;
int inf=(1<<25);
int vis[maxn];
int dis[maxn];
int exf[maxn];
int pre[maxn];
int last[maxn];
int znx;
int Aimee;
bool spfa(){
	memset(vis,0,sizeof(vis));
	memset(dis,0x7f,sizeof(dis));
	memset(exf,0x7f,sizeof(exf));
	while(!q.empty()){
		q.pop();
	}
	q.push(s);
	dis[s]=0;vis[s]=1,pre[t]=-1;
	while(!q.empty()){
		int x=q.front();
		vis[x]=0;
		q.pop();
		for(int i=head[x];i;i=ed[i].ne){
			if(ed[i].v>0&&dis[ed[i].to]>dis[x]+ed[i].c){
				dis[ed[i].to]=dis[x]+ed[i].c;
				pre[ed[i].to]=i;
				exf[ed[i].to]=min(exf[x],ed[i].v);
				if(!vis[ed[i].to]){
					q.push(ed[i].to);
					vis[ed[i].to]=1;
				}
			}
		}
	}
	return pre[t]!=-1;
}
void up(){
	znx+=exf[t];
	Aimee+=exf[t]*dis[t];
	int now=t;
	while(now!=s){
		ed[pre[now]].v-=exf[t];
		ed[pre[now]^1].v+=exf[t];
		now=ed[pre[now]^1].to;
	}
	return ;
} 
int main(){
	scanf("%d%d%d%d",&n,&m,&s,&t);
	for(int i=1;i<=m;++i){
		scanf("%d%d%d%d",&u,&v,&w,&c);
		add(u,v,w,c);
	}
	while(spfa()){
		up();
	}
	cout<<znx<<" "<<Aimee;
	return 0;
}