(简单) POJ 1511 Invitation Cards，SPFA。

　　Description

　　In the age of television, not many people attend theater performances. Antique Comedians of Malidinesia are aware of this fact. They want to propagate theater and, most of all, Antique Comedies. They have printed invitation cards with all the necessary information and with the programme. A lot of students were hired to distribute these invitations among the people. Each student volunteer has assigned exactly one bus stop and he or she stays there the whole day and gives invitation to people travelling by bus. A special course was taken where students learned how to influence people and what is the difference between influencing and robbery.

　　The transport system is very special: all lines are unidirectional and connect exactly two stops. Buses leave the originating stop with passangers each half an hour. After reaching the destination stop they return empty to the originating stop, where they wait until the next full half an hour, e.g. X:00 or X:30, where 'X' denotes the hour. The fee for transport between two stops is given by special tables and is payable on the spot. The lines are planned in such a way, that each round trip (i.e. a journey starting and finishing at the same stop) passes through a Central Checkpoint Stop (CCS) where each passenger has to pass a thorough check including body scan.

　　All the ACM student members leave the CCS each morning. Each volunteer is to move to one predetermined stop to invite passengers. There are as many volunteers as stops. At the end of the day, all students travel back to CCS. You are to write a computer program that helps ACM to minimize the amount of money to pay every day for the transport of their employees.

　　题目是最短路问题，求源点到所有点的来回的最短距离之和，只需要反向边图+正向边图即可。。。

代码如下：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
    
using namespace std;

const int INF = 10e8;
const int MaxN = 1000010;

struct Edge
{
    int v, cost,next;
};

Edge E1[MaxN],E2[MaxN],*E;
int head1[MaxN],head2[MaxN],Ecou1,Ecou2;
int *head;
bool vis[MaxN];
int couNode[MaxN];

bool SPFA(long long lowcost[], int n, int start,int type)
{
    if(type)
        E=E1,head=head1;
    else
        E=E2,head=head2;

    queue <int> que;
    int u, v, c;
    int len;

    for (int i = 1; i <= n; ++i) { lowcost[i] = INF; vis[i] = 0; couNode[i] = 0; }

    vis[start] = 1; lowcost[start] = 0; couNode[start] = 1;
    que.push(start);

    while (!que.empty())
    {
        u = que.front(); que.pop();

        vis[u] = 0;

        for (int i = head[u]; i!=-1; i=E[i].next)
        {
            v = E[i].v; c = E[i].cost;

            if (lowcost[v]>lowcost[u] + c)
            {
                lowcost[v] = lowcost[u] + c;

                if (!vis[v])
                {
                    vis[v] = 1; ++couNode[v]; que.push(v);

                    if (couNode[v]>n) return 0;
                }
            }
        }
    }

    return 1;
}

inline void addEdge(int u, int v, int c,int type)
{
    int *Ecou;
    if(type)
    {
        E=E1,head=head1;
        Ecou=&Ecou1;
    }
    else
    {
        E=E2,head=head2;
        Ecou=&Ecou2;
    }

    E[*Ecou].v=v;
    E[*Ecou].cost=c;
    E[*Ecou].next=head[u];
    head[u]=(*Ecou)++;
}

void init(int N)
{
    Ecou1=Ecou2=0;

    for(int i=1;i<=N;++i)
        head1[i]=head2[i]=-1;
}

long long ans1[MaxN],ans2[MaxN];

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

    int T;
    int N,M;
    int a,b,c;
    long long ans;

    cin>>T;

    while(T--)
    {
        scanf("%d %d",&N,&M);
        init(N);

        for(int i=1;i<=M;++i)
        {
            scanf("%d %d %d",&a,&b,&c);
            addEdge(a,b,c,1);
            addEdge(b,a,c,0);
        }

        SPFA(ans1,N,1,1);
        SPFA(ans2,N,1,0);

        ans=0;
        for(int i=2;i<=N;++i)
            ans+=ans1[i]+ans2[i];

        cout<<ans<<endl;
    }
    
    return 0;
}