hdu5441

小程序：扫一扫查出行
【扫一扫了解最新限行尾号】
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Travel

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 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 m bidirectional 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

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=;
const int maxm=;
const int maxq=;
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[];
long long ans[];
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]=;
}
else {
father[y]=x;
if(rak[x]==rak[y])
++rak[x];
num[x]+=num[y];
num[y]=;
}
}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=;i<=n;i++){
father[i]=i;
num[i]=;
rak[i]=;
} for(int i=;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=;i<q;i++){
scanf("%d",&d);
que[i]=node2(d,i);
}
sort(que,que+q,cmp2);
memset(ans,,sizeof(ans));
tnum=;
for(int i=,j=;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=;i<q;i++)
printf("%lld\n",ans[i]*); }
return ;
}