TZOJ 4712 Double Shortest Paths(最小费用最大流)

描述

Alice and Bob are walking in an ancient maze with a lot of caves and one-way passages connecting them. They want to go from cave 1 to cave n. All the passages are difficult to pass. Passages are too small for two people to walk through simultaneously, and crossing a passage can make it even more difficult to pass for the next person. We define di as the difficulty of crossing passage i for the first time, and ai as the additional difficulty for the second time (e.g. the second person's difficulty is di+ai).
Your task is to find two (possibly identical) routes for Alice and Bob, so that their total difficulty is minimized.

For example, in figure 1, the best solution is 1->2->4 for both Alice and Bob, but in figure 2, it's better to use 1->2->4 for Alice and 1->3->4 for Bob.

输入

There will be at most 200 test cases. Each case begins with two integers n, m (1<=n<=500, 1<=m<=2000), the number of caves and passages. Each of the following m lines contains four integers u, v, di and ai (1<=u,v<=n, 1<=di<=1000, 0<=ai<=1000). Note that there can be multiple passages connecting the same pair of caves, and even passages connecting a cave and itself.

输出

For each test case, print the case number and the minimal total difficulty.

样例输入

4 4
1 2 5 1
2 4 6 0
1 3 4 0
3 4 9 1
4 4
1 2 5 10
2 4 6 10
1 3 4 10
3 4 9 10

样例输出

Case 1: 23
Case 2: 24

题意

找两条从1到N的路使得总花费最小，一条路第一次走花费d，第二次走花费d+a

题解

每个点建两条边一条流量1花费d，一条流量1花费d+a

源点S=0，和1连一条边流量2花费0

汇点T=n+1，和n连一条边流量2花费0

然后跑一边最小费用最大流

代码

#include<bits/stdc++.h>
using namespace std;

const int N=1e5+5;
const int M=2e5+5;
const int INF=0x3f3f3f3f;

int FIR[N],FROM[M],TO[M],CAP[M],FLOW[M],COST[M],NEXT[M],tote;
int pre[N],dist[N],q[400000];
bool vis[N];
int n,m,S,T;
void init()
{
    tote=0;
    memset(FIR,-1,sizeof(FIR));
}
void addEdge(int u,int v,int cap,int cost)
{
    FROM[tote]=u;
    TO[tote]=v;
    CAP[tote]=cap;
    FLOW[tote]=0;
    COST[tote]=cost;
    NEXT[tote]=FIR[u];
    FIR[u]=tote++;

    FROM[tote]=v;
    TO[tote]=u;
    CAP[tote]=0;
    FLOW[tote]=0;
    COST[tote]=-cost;
    NEXT[tote]=FIR[v];
    FIR[v]=tote++;
}
bool SPFA(int s, int t)
{
    memset(dist,INF,sizeof(dist));
    memset(vis,false,sizeof(vis));
    memset(pre,-1,sizeof(pre));
    dist[s] = 0;vis[s]=true;q[1]=s;
    int head=0,tail=1;
    while(head!=tail)
    {
        int u=q[++head];vis[u]=false;
        for(int v=FIR[u];v!=-1;v=NEXT[v])
        {
            if(dist[TO[v]]>dist[u]+COST[v]&&CAP[v]>FLOW[v])
            {
                dist[TO[v]]=dist[u]+COST[v];
                pre[TO[v]]=v;
                if(!vis[TO[v]])
                {
                    vis[TO[v]] = true;
                    q[++tail]=TO[v];
                }
            }
        }
    }
    return pre[t]!=-1;
}
void MCMF(int s, int t, int &cost, int &flow)
{
    flow=0;
    cost=0;
    while(SPFA(s,t))
    {
        int Min = INF;
        for(int v=pre[t];v!=-1;v=pre[TO[v^1]])
            Min = min(Min, CAP[v]-FLOW[v]);
        for(int v=pre[t];v!=-1;v=pre[TO[v^1]])
        {
            FLOW[v]+=Min;
            FLOW[v^1]-=Min;
            cost+=COST[v]*Min;
        }
        flow+=Min;
    }
}
int main()
{
    int ca=0;
    while(scanf("%d%d",&n,&m)!= EOF)
    {
        init();
        for(int i=0,u,v,d,a;i<m;i++)
        {
            scanf("%d%d%d%d",&u,&v,&d,&a);
            addEdge(u,v,1,d);
            addEdge(u,v,1,d+a);
        }
        S=0,T=n+1;
        addEdge(S,1,2,0);
        addEdge(n,T,2,0);
        int cost,flow;
        MCMF(S,T,cost,flow);
        printf("Case %d: %d
",++ca,cost);
    }
    return 0;
}