USACO Section 3.2 Sweet Butter - 还是SPFA好使..

   这道题就是先找出两两间的最短路...然后扫描所有点找出到达所要求点距离和最小的..

   开始我用的Floyd..超时很严重阿.然后想写Bellman-Ford...但发现效率也不高...那就果断SPFA了...果然高效...水过...

Progarm:

/*  
ID: zzyzzy12   
LANG: C++   
TASK: butter
*/      
#include<iostream>      
#include<istream>  
#include<stdio.h>      
#include<string.h>      
#include<math.h>      
#include<stack>  
#include<algorithm>      
#include<queue>   
#define oo 1000000000  
#define ll long long   
using namespace std;   
struct node
{
      int x,y,w,next; 
}line[3000];
int arc[810][810],n,p,c,s[810],_link[810];
queue<int> myqueue;
int SPFA()
{ 
      int i,j,m,h,g,k=oo;
      bool had[810];
      memset(had,false,sizeof(had));
      for (i=1;i<=p;i++)
      { 
             while (!myqueue.empty()) myqueue.pop();
             myqueue.push(i);
             while (!myqueue.empty())
             {
                    h=myqueue.front(); 
                    myqueue.pop();
                    had[h]=false;
                    g=_link[h];
                    while (g)
                    {
                           if (arc[i][line[g].y]>arc[i][line[g].x]+line[g].w)  
                           {  
                                 arc[i][line[g].y]=arc[i][line[g].x]+line[g].w;
                                 if (!had[line[g].y])
                                 {
                                       myqueue.push(line[g].y);
                                       had[line[g].y]=true;
                                 }                                                
                           }
                           g=line[g].next;      
                    }
             }
             m=0;
             for (j=1;j<=n;j++)
               if (arc[i][s[j]]!=oo) m+=arc[i][s[j]];
                 else goto A;
             if (m<k) k=m;
             A: ;           
      } 
      return k; 
}
int main()  
{  
      freopen("butter.in","r",stdin);    
      freopen("butter.out","w",stdout);    
      scanf("%d%d%d",&n,&p,&c);
      int i,j,k,m;
      for (i=1;i<=p;i++) 
         for (j=1;j<=p;j++) arc[i][j]=oo;
      for (i=1;i<=p;i++) arc[i][i]=0;
      memset(_link,0,sizeof(_link));
      for (i=1;i<=n;i++) scanf("%d",&s[i]);
      for (i=1;i<=c;i++)
      {
             scanf("%d%d%d",&j,&k,&m); 
             line[i*2-1].x=j; line[i*2-1].y=k; line[i*2-1].w=m;
             line[i*2-1].next=_link[j]; _link[j]=i*2-1;
             line[i*2].x=k;   line[i*2].y=j;   line[i*2].w=m;
             line[i*2].next=_link[k];  _link[k]=i*2;
      } 
      printf("%d\n",SPFA());
      return 0;     
}