题目链接

Update：我好像刚知道多路增广就是zkw费用流。。

//1314ms	2.66MB 本题优化明显

#include <queue>

#include <cstdio>

#include <cctype>

#include <cstring>

#include <algorithm>

//#define gc() getchar()

#define MAXIN 350000

#define gc() (SS==TT&&(TT=(SS=IN)+fread(IN,1,MAXIN,stdin),SS==TT)?EOF:*SS++)

const int N=5005,M=1e5+5,INF=0x3f3f3f3f;

int n,m,src,des,Enum,cur[N],H[N],nxt[M],to[M],cap[M],cost[M],dis[N];

long long Cost;

bool vis[N];

std::queue<int> q;

char IN[MAXIN],*SS=IN,*TT=IN;

inline int read()

{

	int now=0;register char c=gc();

	for(;!isdigit(c);c=gc());

	for(;isdigit(c);now=now*10+c-'0',c=gc());

	return now;

}

inline void AddEdge(int c,int w,int v,int u)

{

	to[++Enum]=v, nxt[Enum]=H[u], H[u]=Enum, cap[Enum]=w, cost[Enum]=c;

	to[++Enum]=u, nxt[Enum]=H[v], H[v]=Enum, cap[Enum]=0, cost[Enum]=-c;

}

bool SPFA()

{

	memset(vis,0,sizeof vis);

	memset(dis,0x3f,sizeof dis);

	dis[src]=0, q.push(src);

	while(!q.empty())

	{

		int x=q.front();

		q.pop(), vis[x]=0;

		for(int i=H[x],v; i; i=nxt[i])

			if(cap[i] && dis[v=to[i]]>dis[x]+cost[i])

				dis[v]=dis[x]+cost[i], !vis[v]&&(q.push(v),vis[v]=1);

	}

	return dis[des]<INF;

}

int DFS(int x,int f)

{

	if(x==des) return f;

	vis[x]=1;

	for(int &i=cur[x],tmp; i; i=nxt[i])

		if(!vis[to[i]] && cap[i] && dis[to[i]]==dis[x]+cost[i])

			if(tmp=DFS(to[i],std::min(f,cap[i])))

				return cap[i]-=tmp,cap[i^1]+=tmp,Cost+=1ll*tmp*cost[i],tmp;

	return 0;

}

void MCMF()

{

	long long flow=0,tmp;

	while(SPFA())

	{

		for(int i=1; i<=n; ++i) cur[i]=H[i];

		while(tmp=DFS(src,INF)) flow+=tmp;

	}

	printf("%lld %lld\n",flow,Cost);

}

int main()

{

	Enum=1,n=read(),m=read(),src=read(),des=read();

	for(int i=1; i<=m; ++i) AddEdge(read(),read(),read(),read());

	MCMF();

	return 0;

}