Luogp3381 [TEMPLATE] The maximum minimum fee (dijstra fee stream) and p3381dijstra

Description

For example, a network diagram and its source and sink points are provided. The maximum traffic and unit traffic fee are known for each edge, and the maximum network flow and the minimum fee are obtained.

Input/Output Format

Input Format:

The first line contains four positive integers N, M, S, and T, indicating the number of vertices, the number of directed edges, The Source Vertex sequence number, and the sink vertex sequence number.

Each row in the next M line contains four positive integers, ui, vi, wi, and fi, indicating that the I-th directed edge departs from the ui and reaches vi, the edge weight is wi (that is, the maximum traffic for this edge is wi), and the unit traffic fee is fi.

Output Format:

A row contains two integers, which are the maximum traffic and the minimum fee in the case of the maximum traffic.

Input and Output sample input sample #1: Copy

4 5 4 34 2 30 24 3 20 32 3 20 12 1 30 91 3 40 5

Output example #1: Copy

50 280

Description

Time-Space limit: 1000 ms, 128 M

(BYX: The last two points are changed to 1200 ms)

Data scale:

For 30% of data: N <= 10, M <= 10

For 70% of data: N <= 1000, M <= 1000

For 100% of data: N <= 5000, M <= 50000

Example:

, The optimal solution is as follows:

The first stream is 4 --> 3, the traffic is 20, and the fee is 3*20 = 60.

The second stream is 4 --> 2 --> 3, the traffic is 20, and the fee is (2 + 1) * 20 = 60.

The third stream is 4 --> 2 --> 1 --> 3, the traffic is 10, and the fee is (2 + 9 + 5) * 10 = 160.

Therefore, the maximum traffic is 50. In this case, the minimum fee is 60 + 60 + 160 = 280.

Therefore, the output is 50 280.

**The dijstra fee stream is really not very fast**

**SPFA**

**I have a good blog.**

**Http://www.yhzq-blog.cc/%E6%9C%80%E5%B0%8F%E8%B4%B9%E7%94%A8%E6%9C%80%E5%A4%A7%E6%B5%81%E7% AE %97%E6%B3%95%E6%80%BB%E7%BB%93/**

**The last sentence is * h, not * dis.**

**In my personal understanding, because h exists during the most short circuit, the dis here is not actually dis, and h is actually dis.**

// luogu-judger-enable-o2#include<cstdio>#include<cstring>#include<queue>#define Pair pair<int,int>#define fi first#define se second#define AddEdge(x,y,f,z) add_edge(x,y,f,z);add_edge(y,x,0,-z);#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,MAXN,stdin),p1==p2)?EOF:*p1++)char buf[1<<20],*p1=buf,*p2=buf;using namespace std;const int MAXN=1e6+1,INF=1e8+10;inline int read(){ char c=getchar();int x=0,f=1; while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();} return x*f;}int N,M,S,T;struct node{ int u,v,f,w,nxt;}edge[MAXN];int head[MAXN],num=2;inline void add_edge(int x,int y,int f,int z){ edge[num].u=x; edge[num].v=y; edge[num].f=f; edge[num].w=z; edge[num].nxt=head[x]; head[x]=num++;}int h[MAXN],dis[MAXN],PrePoint[MAXN],PreEdge[MAXN];Pair Dij(){ int ansflow=0,anscost=0; while(1) { priority_queue<Pair>q; memset(dis,0xf,sizeof(dis)); dis[S]=0; q.push(make_pair(0,S)); while(q.size()!=0) { Pair p=q.top();q.pop(); if(-p.fi!=dis[p.se]) continue; if(p.se==T) break; for(int i=head[p.se];i!=-1;i=edge[i].nxt) { int nowcost=edge[i].w+h[p.se]-h[edge[i].v]; if(edge[i].f>0&&dis[edge[i].v]>dis[p.se]+nowcost) { dis[edge[i].v]=dis[p.se]+nowcost; q.push(make_pair(-dis[edge[i].v],edge[i].v)); PrePoint[edge[i].v]=p.se; PreEdge[edge[i].v]=i; } } } if(dis[T]>INF) break; for(int i=0;i<=N;i++) h[i]+=dis[i]; int nowflow=INF; for(int now=T;now!=S;now=PrePoint[now]) nowflow=min(nowflow,edge[PreEdge[now]].f); for(int now=T;now!=S;now=PrePoint[now]) edge[PreEdge[now]].f-=nowflow, edge[PreEdge[now]^1].f+=nowflow; ansflow+=nowflow; anscost+=nowflow*h[T]; } return make_pair(ansflow,anscost);}int main(){ #ifdef WIN32 freopen("a.in","r",stdin); #endif memset(head,-1,sizeof(head)); N=read(),M=read(),S=read(),T=read(); for(int i=1;i<=M;i++) { int x=read(),y=read(),f=read(),z=read(); AddEdge(x,y,f,z); } Pair ans=Dij(); printf("%d %d",ans.fi,ans.se); return 0;}