hdu 5441

Travel

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 4685    Accepted Submission(s): 1535

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 n cities and 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 x minutes, how many pairs of city (a,b) are there that Jack can travel from city a to b without going berserk?

 

 

Input

The first line contains one integer T,T≤5, which represents the number of test case.

For each test case, the first line consists of three integers n,m and q where n≤20000,m≤100000,q≤5000. The Undirected Kingdom has n cities and mbidirectional roads, and there are q queries.

Each of the following m lines consists of three integers a,b and d where a,b∈{1,...,n} and d≤100000. It takes Jack d minutes to travel from city a to city b and vice versa.

Then q lines follow. Each of them is a query consisting of an integer x where x is the time limit before Jack goes berserk.

 

 

Output

You should print q lines for each test case. Each of them contains one integer as the number of pair of cities (a,b) which Jack may travel from a to b within the time limit x.

Note that (a,b) and (b,a) are counted as different pairs and a and 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

 

 

Source

2015 ACM/ICPC Asia Regional Changchun Online

 

 

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#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<iostream>
#include<vector>
using namespace std;
typedef long long ll;
#define N 100009
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define mem(a,b) memset(a,b,sizeof(a))
int  t,n,m,q;
struct Node{
    int u,v,w;
}nod[N];
struct Nod{
    int id,w;
}no[N];
int pre[20009];
int cnt[20009];
bool cmp(Node a,Node b){
    return a.w<b.w;
}
bool cmp1(Nod a,Nod b){
    return a.w<b.w;
}
void init()
{
    for(int i=0;i<=n;i++){
        pre[i]=i;
        cnt[i]=1;
    }
}
int find(int x)
{
    return pre[x]=x==pre[x]?x:find(pre[x]);
}
void unite(int u,int v)
{  /*
    pre[v]=u;
    cnt[u]+=cnt[v];
    */
    pre[u]=v;
    cnt[v]+=cnt[u];
    //上面两种写法都是对的。
}
int ret[N];
int  main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&n,&m,&q);
        init();
        for(int i=0;i<m;i++){
            scanf("%d%d%d",&nod[i].u,&nod[i].v,&nod[i].w);
            }
            sort(nod,nod+m,cmp);
            for(int i=0;i<q;i++){
                scanf("%d",&no[i].w);
                no[i].id=i;
            }
            sort(no,no+q,cmp1);
            int j=0,ans=0;
            int u,v;
            //预处理
            for(int i=0;i<q;i++){
                while(j<m&&nod[j].w<=no[i].w){
                    u=find(nod[j].u);
                    v=find(nod[j].v);
                    if(u!=v){
                        int tmp=cnt[v]+cnt[u];
                        //任意一对点都行
                        ans+=(tmp*(tmp-1))-(cnt[u]*(cnt[u]-1))-(cnt[v]*(cnt[v]-1));
                        //减去已经加过的
                        unite(u,v);
                    }
                    j++;
                }
                ret[no[i].id]=ans;
            }            
            for(int i=0;i<q;i++){
                printf("%d\n",ret[i]);
            }        
    }
    return 0;
}