hdu5431 Travel

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 <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
const int oo = 0x3f3f3f3f;
const int N = 200010;
const int M = 100100;
typedef long long LL;
int father[N], num[N], QQ[N], n, m;
/**< num[i]表示i这个根节点 所包含的节点个数*/
struct da
{
    int u, v, w;
    bool operator < (const da &a)const
    {
        return w < a.w;
    }
}as[N];
struct da1
{
    int id, x;
    bool operator < (const da1 &a)const
    {
        return x < a.x;
    }
}ac[N];
int fin(int x)
{
    if(x != father[x])
        father[x] = fin(father[x]);
    return father[x];
}
void init()
{
    for(int i = 0; i <= n; i++)
    {
        father[i] = i;
        num[i] = 1;
    }
    memset(QQ, 0, sizeof(QQ));
}
int main()
{
    int T, i, j, k;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d %d %d", &n, &m, &k);
        init();
        for(i = 0; i < m; i++)
            scanf("%d %d %d", &as[i].u, &as[i].v, &as[i].w);
        sort(as, as+m);
        for(i = 0; i < k; i++)
        {
             scanf("%d", &ac[i].x);
             ac[i].id = i;
        }
        sort(ac, ac+k);
        int ans=0;j = 0;
        for(i = 0; i < k; i++)
        {
            while(j < m && as[j].w <= ac[i].x)
            {
                int x = fin(as[j].u);
                int y = fin(as[j].v);
                j++;
                if(x != y)
                {
                    father[x] = y;
                    ans += num[y]*num[x];/**< 集合合并 */
                    num[y] += num[x];
                }
            }
            QQ[ac[i].id] = 2*ans;
        }
        for(i = 0; i < k; i++)
            printf("%d\n", QQ[i]);

    }
    return 0;
}