hdu5441

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1225    Accepted Submission(s): 443

Problem Description

Jack likes to travel around the world, but he doesn’t like to wait. Now, he is traveling in the Undirected Kingdom. There are m bidirectional roads connecting the cities. Jack hates waiting too long on the bus, but he can rest at every city. Jack can only stand staying on the bus for a limited time and will go berserk after that. Assuming you know the time it takes to go from one city to another and that the time Jack can stand staying on a bus is b without going berserk?

 

 

Input

The first line contains one integer x is the time limit before Jack goes berserk.

 

 

Output

You should print b must be different cities.

 

 

Sample Input

1 5 5 3 2 3 6334 1 5 15724 3 5 5705 4 3 12382 1 3 21726 6000 10000 13000

 

 

Sample Output

2 6 12

 

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=20010;
const int maxm=100010;
const int maxq=5005;
struct node1{
   int u,v,w;
   node1(){}
   node1(int u,int v,int w):u(u),v(v),w(w) {}
}g[maxm];

struct node2{
   int d,id;
   node2(){}
   node2(int d,int id):d(d),id(id) {}
}que[5005];
long long  ans[5005];
int num[maxn], rak[maxn],father[maxn];
bool cmp1(struct node1 t1,struct node1 t2){
      return t1.w<t2.w;
}
bool cmp2(struct node2 t1,struct node2 t2){
   return t1.d<t2.d;
}


int find(int x){
   if(x!=father[x])
       father[x]=find(father[x]);
    return father[x];
}
long long  tnum;
void Union(int u,int v){
    int x=find(u);
    int y=find(v);
    if(x==y)
        return ;
    tnum+=num[x]*num[y];

    if(rak[x]<rak[y]){
           father[x]=y;
           num[y]+=num[x];
           num[x]=0;
    }
    else {
        father[y]=x;
       if(rak[x]==rak[y])
            ++rak[x];
       num[x]+=num[y];
       num[y]=0;
    }
}

int main(){
   int t;
   scanf("%d",&t);
   while(t--){
      int n,m,q;
    
      scanf("%d%d%d",&n,&m,&q);
      int u,v,w;

    for(int i=1;i<=n;i++){
      father[i]=i;
      num[i]=1;
      rak[i]=0;
    }

      for(int i=0;i<m;i++){
          scanf("%d%d%d",&u,&v,&w);
           g[i]=node1(u,v,w);
      }
      sort(g,g+m,cmp1);
      int d;
      for(int i=0;i<q;i++){
        scanf("%d",&d);
        que[i]=node2(d,i);
      }
      sort(que,que+q,cmp2);
        memset(ans,0,sizeof(ans));
      tnum=0;
      for(int i=0,j=0;i<q;i++){
          int cur=que[i].d;
          while(j<m){
            node1 temp=g[j];
    
             if(cur>=temp.w){
            
                Union(temp.u,temp.v);
             }
             else
                 break;
             j++;
          }
         ans[que[i].id]=tnum;
      }
      for(int i=0;i<q;i++)
          printf("%lld\n",ans[i]*2);

   }
   return 0;
}