USCAO3.26Sweet Butter（SPFA)

最短路复杂度估计错误 以为SPFA是N*m的 用了dij超时 用SPFA直接跑就好了

O（k*e) K 一般为2,3；

 /*
     ID: shangca2
     LANG: C++
     TASK: butter
  */
 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<queue>
 using namespace std;
 #define M 1500
 #define P 850
 #define N 550
 #define INF 0xfffffff
 struct node
 {
     int u,v,w,next;
 }ed[M<<];
 int t,head[P],dis[P],a[N],n,p,c,vis[P],f[P],k[P],o[P][P];
 void init()
 {
     t = ;
     memset(head,-,sizeof(head));
 }
 void add(int u,int v,int w)
 {
     ed[t].u = u;
     ed[t].v = v;
     ed[t].w = w;
     ed[t].next = head[u];
     head[u] = t++;
 }
 void spfa(int st)
 {
     int i;
     memset(vis,,sizeof(vis));
     for(i = ; i <= p ; i++)
     dis[i] = INF;
     dis[st] = ;
     queue<int>q;
     vis[st] = ;
     q.push(st);
     while(!q.empty())
     {
         int u = q.front();
         q.pop();
         vis[u] = ;
         for(i = head[u]; i != - ;i = ed[i].next)
         {
             int v = ed[i].v;
             if(ed[i].w+dis[u]<dis[v])
             {
                 dis[v] = ed[i].w+dis[u];
                 if(!vis[v])
                 {
                     vis[v] = ;
                     q.push(v);
                 }
             }
         }
     }
     for(i = ; i <= p ; i++)
     {
         if(o[st][i]>dis[i])
         {
             o[st][i] = dis[i];
             o[i][st] = dis[i];
         }
     }
 }
 int main()
 {
     freopen("butter.in","r",stdin);
     freopen("butter.out","w",stdout);
     int i,j,u,v,w;
     init();
     cin>>n>>p>>c;
     for(i  =  ;i <= p ; i++)
         for(j = ; j <= p ; j++)
         o[i][j] = INF;
     for(i = ; i <= n ;i++)
         cin>>a[i];
     for(i = ; i <= c ;i++)
     {
         scanf("%d%d%d",&u,&v,&w);
         add(u,v,w);
         add(v,u,w);
     }
     for(i = ; i <= n ;i++)
     {
         if(!f[a[i]])
         {
             f[a[i]] = ;
             spfa(a[i]);
         }
     }
     int minz = INF;
     for(i = ; i <= p ; i++)
     {
         int s = ;
         for(j =  ; j <= n ; j++)
         {
             if(a[j]!=i)
             s+=o[a[j]][i];
         }
         minz = min(minz,s);
     }
     printf("%d\n",minz);
     return ;
 }