POJ-1511 Invitation Cards （双向单源最短路）

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：单向边，求从起点1到每个点的最短路然后再回到起点1的最短路之和。

题目解析：朴素算法TLE。求每个点到起点的最短时，将每一条边倒过来，又转化成了起点1到每一个点的距离之和。（逆向思维）

 

代码如下：

 # include<iostream>
 # include<cstdio>
 # include<cstring>
 # include<queue>
 # include<algorithm>
 using namespace std;
 const int N=;
 const long long INF=<<;
 struct edge
 {
     int to,w,nxt;
 };
 edge e[N+];
 int n,cnt,head[N+],a1[N+],a2[N+],c[N+];
 long long dis[N+];
 void add(int u,int v,int w)
 {
     e[cnt].to=v;
     e[cnt].w=w;
     e[cnt].nxt=head[u];
     head[u]=cnt++;
 }
 long long spfa()
 {
     fill(dis,dis+n+,INF);
     queue<int>q;
     q.push();
     dis[]=;
     while(!q.empty())
     {
         int u=q.front();
         q.pop();
         for(int i=head[u];i!=-;i=e[i].nxt){
             if(dis[e[i].to]>dis[u]+e[i].w){
                 dis[e[i].to]=dis[u]+e[i].w;
                 q.push(e[i].to);
             }
         }
     }
     long long res=;
     for(int i=;i<=n;++i)
         res+=dis[i];
     return res;
 }
 int main()
 {
     int T,m;
     scanf("%d",&T);
     while(T--)
     {
         cnt=;
         scanf("%d%d",&n,&m);
         fill(head,head+n+,-);
         for(int i=;i<m;++i){
             scanf("%d%d%d",&a1[i],&a2[i],&c[i]);
             add(a1[i],a2[i],c[i]);
         }
         long long ans=spfa();
         //cout<<ans<<endl;
         cnt=;
         fill(head,head+n+,-);
         for(int i=;i<m;++i)
             add(a2[i],a1[i],c[i]);
         ans+=spfa();
         printf("%lld\n",ans);
     }
     return ;
 }