HDU 5441 Travel

Travel

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 94    Accepted Submission(s): 32

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 band 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

 

并查集离线

/* ***********************************************
Author        :guanjun
Created Time  :2015/9/13 15:57:19
File Name     :4.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 100000+10
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;
struct node{
    int x,y,z;
}edge[maxn];
bool cmp(node a,node b){
    return a.z<b.z;
}
struct pk{
    int w,id;
}q[maxn];
bool cmp1(pk a,pk b){
    return a.w<b.w;
}
int ans[maxn];
int fa[maxn],num[maxn];
int findfa(int x){
    if(x==fa[x])return fa[x];
    else return fa[x]=findfa(fa[x]);
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    int n,m,Q,t;
    cin>>t;
    while(t--){
        cin>>n>>m>>Q;
        for(int i=0;i<m;i++)scanf("%d%d%d",&edge[i].x,&edge[i].y,&edge[i].z);
        for(int i=0;i<Q;i++)scanf("%d",&q[i].w),q[i].id=i;
        sort(edge,edge+m,cmp);
        sort(q,q+Q,cmp1);
        for(int i=1;i<=n;i++)fa[i]=i,num[i]=1;
        int cnt=0,j=0;
        for(int i=0;i<Q;i++){
            while(j<m&&edge[j].z<=q[i].w){
                int x=findfa(edge[j].x);
                int y=findfa(edge[j].y);
                if(x!=y){
                    //当 联通块加入一个符合要求的点的时候相应的应该
                    //增加的方案数
                    cnt+=num[x]*num[y];
                    fa[x]=y;
                    num[y]+=num[x];
                }
                j++;
            }
            ans[q[i].id]=2*cnt;
        }
        for(int i=0;i<Q;i++){
            printf("%d
",ans[i]);
        }
    }
    return 0;
}