POJ1511

                                                                                         Invitation Cards

Time Limit: 8000MS		Memory Limit: 262144K
Total Submissions: 14928		Accepted: 4831

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

 

 

 

思路：这题就是要求从点1到其他各个点，然后从其他各个点回到点1的最小总距离。很明显，首先会想到DIJ,但是这题用DIJ速度太慢了。因为DIJ的复杂度是O(N^2)的。用FLOYD也不行，因为FLOYD需要使用邻接矩阵，但是这而N太大了。不能使用建立这么大的邻接矩阵。然后就剩下Bellman-fordl和SPFA了。这个里面Bellman-ford的复杂度明显比SPFA高。所以，我们使用SPFA+邻接表+逆邻接表解决这个问题。在求其他点到点1的距离时，使用逆邻接表存储，只要SPFA就可以求出来了。而如果不这样，就需要每一次，求出点X(X!=1)到其他点的距离，然后选择到点1的距离。不断这样的循环，需要N-1次。而使用逆邻接表只需要一次就可以了。另外：注意  本题不要使用STL否则会超时。本题邻接和逆邻接链表都需要使用静态的。

 

 

#include <cstdlib>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>



#define MAXCOST 1000000001
#define MAX_Q_SIZE 1000000+4


using namespace std;

struct Node{
       int v;
       int cost;
       int next;
       
       };



int d[1000000+4];


       
Node node[2000000+4];
int begin,end;
int vt[1000000+4];

int revt[1000000+4];


      
      



template <typename first_type>
class queue{
      first_type q[MAX_Q_SIZE];
      int begin,end;
      int maxsize;
      int cursize;
      
      
      public :
             
      queue()
      {begin=end=0;maxsize=MAX_Q_SIZE-1;cursize=0;}



     int empty()
     {if(cursize<=0)
      return 1;
      return 0;
     }
     
     int clear()
     {begin=end=0;maxsize=MAX_Q_SIZE-1;cursize=0;return 1;}
     
     
     int pop()
     {if(cursize<=0)
       return 0;
       begin++;
       cursize--;
       return 1;
     }
     
     int size()
     {if(cursize<=0)
     return 0;
     return cursize;
     }
     
     int push(first_type x)
     {if(cursize>=maxsize)
        return 0;
      q[end++]=x;
      cursize++;
      return 1;
     }
     
     first_type front()
     {
         if(cursize<=0)
         return NULL;
         return q[begin];
     }
     
};



queue<int>q;

      
      
      
      
      
      


int main(int argc, char *argv[])
{
    
    int n;
    
    int i,j,k;
    
    int m;
    
    int t;
    
    scanf("%d",&t);
    
    
    while(t--)
    {
              scanf("%d%d",&n,&m);
              begin=end=0;
              
              for(i=1;i<=n;i++)
              {revt[i]=-1;vt[i]=-1;}
              
              
             
              q.clear();
              for(i=1;i<=m;i++)
              {
                               int a,b;
                               long long cost;
                               scanf("%d%d%d",&a,&b,&cost);
                               Node tmp;
                               tmp.v=b;
                               tmp.cost=cost;
                               
                               
                               int tt=vt[a];
                               tmp.next=tt;
                               node[begin]=tmp;
                               begin++;
                               vt[a]=begin-1;
                               
                               
                               tmp.v=a;
                               tt=revt[b];
                               tmp.next=tt;
                               node[begin]=tmp;
                               begin++;
                               revt[b]=begin-1;
                               
                               
                               
                               
              }
              
              
              
              int lowcost=MAXCOST;
              long long sumcost=0;
             
              
              
              
              
              
                               k=1;
                               for(i=1;i<=n;i++)
                               {d[i]=MAXCOST;}
                               
                               
                               
                               q.clear();
                               int it;
                               for(it=vt[k];it!=-1;it=node[it].next)
                               {d[node[it].v]=node[it].cost;q.push(node[it].v);
                               }
                               
                               
                              
                               
                               while(!q.empty())
                               {
                                       int x=q.front();
                                       q.pop();
                                       for(it=vt[x];it!=-1;it=node[it].next)
                                       {
                                                                            if(d[node[it].v]>d[x]+node[it].cost)
                                                                            {d[node[it].v]=d[x]+node[it].cost;q.push(node[it].v);
                                                                            
                                                                            }
                                       }
                                       
                               }
                               
              
                              
                                       int y;
                                       for(y=2;y<=n;y++)
                                       {sumcost+=d[y];}
                              
                               
                               
                               
                               
                               
                               
                               
              
              
              
                               for(i=1;i<=n;i++)
                               {d[i]=MAXCOST;}
                               
                               
                              
                               q.clear();
                               
                               for(it=revt[k];it!=-1;it=node[it].next)
                               {d[node[it].v]=node[it].cost;
                               q.push(node[it].v);
                               
                               }
                               
                               
                              
                               
                               while(!q.empty())
                               {
                                       int x=q.front();
                                       q.pop();
                                       for(it=revt[x];it!=-1;it=node[it].next)
                                       {
                                                                            if(d[node[it].v]>d[x]+node[it].cost)
                                                                            {d[node[it].v]=d[x]+node[it].cost;
                                                                            q.push(node[it].v);//
                                                                            
                                                                            }
                                       }
                                       
                               }
                               
              
                               
                                       
                                       for(y=2;y<=n;y++)
                                       {sumcost+=d[y];}
                               
                               
                               
                               
                              
                               
              
              
                                               
                                               
                               
                 // printf("%lld\n",sumcost);             
              cout<<sumcost<<endl;
              
    }         
    
    
    
    
    
    
    //system("PAUSE");
    return EXIT_SUCCESS;
}