poj 1511 Invitation Cards

/*
Invitation Cards
Time Limit: 8000MS         Memory Limit: 262144K
Total Submissions: 15101         Accepted: 4897
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.
Input

The input consists of N cases. The first line of the input contains only positive integer N. Then follow the cases. Each case begins with a line containing exactly two integers P and Q, 1 <= P,Q <= 1000000. P is the number of stops including CCS and Q the number of bus lines. Then there are Q lines, each describing one bus line. Each of the lines contains exactly three numbers - the originating stop, the destination stop and the price. The CCS is designated by number 1. Prices are positive integers the sum of which is smaller than 1000000000. You can also assume it is always possible to get from any stop to any other stop.
Output

For each case, print one line containing the minimum amount of money to be paid each day by ACM for the travel costs of its volunteers.
Sample Input

2
2 2
1 2 13
2 1 33
4 6
1 2 10
2 1 60
1 3 20
3 4 10
2 4 5
4 1 50
Sample Output

46
210
Source

Central Europe 1998
//貌似我用dijskla加优先队列优化比我用sfpa快些、看来我的sfpa不会优化呀
求原图最短路径、再求反向最短路径

----摘抄自江财小子
*/
#include <iostream>
#include<algorithm>
#include<queue>
#include<cmath>
#include<string.h>
#include<stdio.h>
#include<stdlib.h>
using namespace std;
#define maxn 1000005
#define max 1000000100
int v[maxn];
bool b[maxn];
__int64 dist[maxn];
int p,q;
struct node
{
    int to;
    int next;
    __int64 value;
} eg[maxn];
struct Recode
{
    int x,y;
    __int64 value;
} rc[maxn];
void spfa()
{
    memset(b,0,sizeof(b));
    queue<int> Q;
    Q.push(1);
    b[1]=1;
    int e,i,k;
    while(!Q.empty())
    {
        i=Q.front();
        Q.pop();
        b[i]=0;
        for(e=v[i];e!=-1;e=eg[e].next)
        {
            k=eg[e].to;
            if(dist[k]>dist[i]+eg[e].value)
            {
                if(!b[k])
                {
                    Q.push(k);
                    b[k]=1;
                }
                dist[k]=dist[i]+eg[e].value;
            }
        }
    }
}
int main()
{
    int t;
    int i;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d",&p,&q);
        for(i=1; i<=p; i++)
            v[i]=-1;
        for(i=0; i<q; i++)
        {
            scanf("%d %d %I64d",&rc[i].x,&rc[i].y,&rc[i].value);
            eg[i].value=rc[i].value;//边权值
            eg[i].to=rc[i].y;//边的顶点(有向边)
            eg[i].next=v[rc[i].x];//边的指向(邻接表的下个狐节点)
            v[rc[i].x]=i;//更新头结点的指向
        }
        for(i=2; i<=q; i++)
            dist[i]=max;
        spfa();
        __int64 ans=0;
        for(i=1; i<=p; i++)
            ans+=dist[i];
        for(i=1; i<=p; i++)
            v[i]=-1;
        for(i=0; i<q; i++)
        {
            eg[i].value=rc[i].value;//边权值
            eg[i].to=rc[i].x;//边的顶点(有向边)
            eg[i].next=v[rc[i].y];//边的指向(邻接表的下个狐节点)
            v[rc[i].y]=i;//更新头结点的指向
        }
        for(i=2; i<=q; i++)
            dist[i]=max;
        spfa();
        for(i=1; i<=p; i++)
            ans+=dist[i];
        printf("%I64d
",ans);
    }
    return 0;
}