Travel

Travel

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1583    Accepted Submission(s): 564

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

 题意：Jack喜欢到处旅游，但是他在不停的行程中确有时间限制，超过时间就会晕车。当然他到达某一个地方会休息。给一他在行程中会晕车的时间限制，问他可以有多少像（a->b这样的旅游路线，当然b->a看做是不同的。输入数据给你n个city，m个路线，q个查询。m个路线，每个路线有x->y，及x->y的行程时间。

Attention：TLE，由时间转化成空间……就是先打表

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

#define maxn 100008

int f[maxn], num[maxn], Harsh[maxn];

struct node
{
    int u, v, d;
}P[maxn];

int found(int x)
{
    if(f[x] != x)
        f[x] = found(f[x]);
    return f[x];
}

int cmp(node a, node b)
{
    return a.d < b.d;
}

int main()
{
    int c, n, m, q, s, maxx;
    scanf("%d", &c);
    while(c--)
    {
        maxx = 0;
        scanf("%d%d%d", &n, &m, &q);
        for(int i = 0; i <= n; i++)
            f[i] = i, num[i] = 1;    // num最开始初始化为1……
        memset(Harsh, 0, sizeof(Harsh));

        for(int i = 0; i < m; i++)
        {
            scanf("%d%d%d", &P[i].u, &P[i].v, &P[i].d);
            maxx = max(maxx, P[i].d);
        }
        sort(P, P+m, cmp);   // 处理前应先排序，
        for(int i = 0; i < m; i++)
        {
            int x = found(P[i].u), y = found(P[i].v);
            if(x != y)   // 如果在一个集合，说明已经算过了，就不用算了
            {
                f[x] = y;
                Harsh[P[i].d] += num[x]*num[y]*2;
                num[y] += num[x];
            }
        }
        for(int i = 1; i <= maxx; i++)
            Harsh[i] += Harsh[i-1];
        while(q--)
        {
            scanf("%d", &s);
            if(s > maxx)
                printf("%d
", Harsh[maxx]);   // 注意
            else
                printf("%d
", Harsh[s]);
        }
    }
    return 0;
}