POJ 1511 Invitation Cards (spfa的邻接表)

Invitation Cards

Time Limit : 16000/8000ms (Java/Other)   Memory Limit : 524288/262144K (Java/Other)

Total Submission(s) : 7   Accepted Submission(s) : 1

Problem 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：给出n个点和n条有向边，求所有点到源点1的来回最短路之和（保证每个点都可以往返源点1）
解题思路：这个数据范围太大，明显的不能用floyd，dijstra，bellman-ford这些算法，用spfa的话也不能用邻接矩阵存，
　　　　　因为点太多了，所以采用spfa的邻接表存储搞定
　　　　　稍微有点注意的地方是，来回之和只需要将所有的边反向再从1到所有点求最短路就是他们的最短回路

AC代码：

 #include <stdio.h>
 #include <string.h>
 #define inf 9999999999
 #include <iostream>
 #include <queue>
 #include <algorithm>
 using namespace std;
 struct node
 {
     int to;
     int w;
     int next;
 };
 queue <int > q;
 int n,m;
 node list[];
 node list1[];
 int vis[];
 int dis[];
 int h1[];
 int h2[];
 void spfa()
 {
     int i,j,u;
     for (i = ; i <= n; i ++)
     {
         dis[i] = inf;
         vis[i] = ;
     }
     q.push();
     dis[] = ;
     vis[] = ;

     while (!q.empty())
     {
         u = q.front();
         q.pop();
         vis[u] = ;
         for (j = h1[u]; j ; j = list[j].next)
         {
             if (dis[list[j].to] > dis[u]+list[j].w)
             {
                 dis[list[j].to] = dis[u]+list[j].w;
                 if (!vis[list[j].to])
                 {
                     q.push(list[j].to);
                     vis[list[j].to] = ;
                 }
             }
         }
     }
 }
 void spfa1()
 {
     int i,j,u;
     for (i = ; i <= n; i ++)
     {
         dis[i] = inf;
         vis[i] = ;
     }
     q.push();
     dis[] = ;
     vis[] = ;

     while (!q.empty())
     {
         u = q.front();
         q.pop();
         vis[u] = ;
         for (j = h2[u]; j ; j = list1[j].next)
         {
             if (dis[list1[j].to] > dis[u]+list1[j].w)
             {
                 dis[list1[j].to] = dis[u]+list1[j].w;
                 if (!vis[list1[j].to])
                 {
                     q.push(list1[j].to);
                     vis[list1[j].to] = ;
                 }
             }
         }
     }
 }
 int main ()
 {
     int i,j,t,u,v,w,ans;
     scanf("%d",&t);
     while (t --)
     {
         scanf("%d%d",&n,&m);
         memset(h1,,sizeof(h1));
         memset(h2,,sizeof(h2));
         for (ans = ,i = ; i < m; i ++)
         {
             scanf("%d%d%d",&u,&v,&w);
             node temp = {v,w,};
             list[ans] = temp;
             list[ans].next = h1[u];
             h1[u] = ans;
             temp.to = u;
             list1[ans] = temp;
             list1[ans].next = h2[v];
             h2[v] = ans;
             ans ++;
         }
         long long sum = ;
         spfa();
         for (i = ; i <= n; i ++)
             sum += dis[i];
         spfa1();
         for (i = ; i <= n; i ++)
             sum += dis[i];
         printf("%lld\n",sum);
     }
     return ;
 }