POJ-1511 Invitation Cards( 最短路，spfa )

题目链接：http://poj.org/problem?id=1511

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

题目大意：给定一个有向图，问从起点到每一个点一个来回共需要多少时间
解题思路：将有向线段的起点和终点互换就可以得到任意点到起点的最短路了，由于数据量太大，需要用spfa，同时用cin也会超时。本来还想写一发dfs的，发现会爆栈，就只贴了bfs的码

 #include<iostream>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 #include<map>
 #include<cstdio>
 #include<queue>

 using namespace std;

 const int LEN = ;
 const int INF = 0x3f3f3f3f;

 struct Edge{
     int to, next;
     long long val;
 }edge[][LEN];

 int h[][LEN], cnt1, cnt2;
 int p, q;
 long long dis[LEN];
 bool vis[LEN];

 void spfa_bfs( int cnt ){
     queue<int> Q;
     memset( dis, INF, LEN * sizeof( long long ) );
     memset( vis, false, LEN * sizeof( bool ) );

     dis[] = ;
     Q.push(  );
     vis[] = true;

     while( !Q.empty() ){
         int x;
         x = Q.front(); Q.pop(); vis[x] = false;

         for( int k = h[cnt][x]; k!= ; k = edge[cnt][k].next ){
             int y = edge[cnt][k].to;
             if( dis[x] + edge[cnt][k].val < dis[y] ){
                 dis[y] = dis[x] + edge[cnt][k].val;
                 if( !vis[y] ){
                     vis[y] = true;
                     Q.push( y );
                 }
             }
         }
     }
 }

 int main(){
     int n;
     scanf( "%d", &n );
     while( n-- ){
         cnt1 = cnt2 = ;
         scanf( "%d%d", &p, &q );
         int beg, end;
         long long val;
         memset( h, , sizeof( h ) );
         for( int i = ; i < q; i++ ){
             scanf( "%d%d%lld", &beg, &end, &val );
             edge[][cnt1].to = end;
             edge[][cnt1].val = val;
             edge[][cnt1].next = h[][beg];
             h[][beg] = cnt1++;

             edge[][cnt2].to = beg;
             edge[][cnt2].val = val;
             edge[][cnt2].next = h[][end];
             h[][end] = cnt2++;
         }

         long long ans = ;
         spfa_bfs(  );
         for( int i = ; i <= p; i++ ) ans += dis[i];
         spfa_bfs(  );
         for( int i = ; i <= p; i++ ) ans += dis[i];

         cout << ans << endl;
     }

     return ;
 }