http://acm.hdu.edu.cn/showproblem.php?pid=3938
Portal
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 820 Accepted Submission(s): 425
Problem Description
ZLGG found a magic theory that the bigger banana the bigger banana peel .This important theory can help him make a portal in our universal. Unfortunately, making a pair of portals will cost min{T} energies. T in a path between point V and point U is the length of the longest edge in the path. There may be lots of paths between two points. Now ZLGG owned L energies and he want to know how many kind of path he could make.
Input
There are multiple test cases. The first line of input contains three integer N, M and Q (1 < N ≤ 10,000, 0 < M ≤ 50,000, 0 < Q ≤ 10,000). N is the number of points, M is the number of edges and Q is the number of queries. Each of the next M lines contains three integers a, b, and c (1 ≤ a, b ≤ N, 0 ≤ c ≤ 10^8) describing an edge connecting the point a and b with cost c. Each of the following Q lines contain a single integer L (0 ≤ L ≤ 10^8).
Output
Output the answer to each query on a separate line.
Sample Input
10 10 10 7 2 1 6 8 3 4 5 8 5 8 2 2 8 9 6 4 5 2 1 5 8 10 5 7 3 7 7 8 8 10 6 1 5 9 1 8 2 7 6
Sample Output
36 13 1 13 36 1 36 2 16 13
题意：给出一个连通图，以及每条路径的长度，定义从u走到v的所有边的最长边作为消耗的能量，q次询问，每次给出一个L能量，问存在多少这样的
分析：很明显是并查集，但是每询问一次就做一次并查集肯定会超时，所以应采用离线算法，先把所有答案全求出来，然后一块输出；首先把边权从小到大排序，然后把询问也从小到大排序，对于每次询问，只加入小于询问的边到一个集合中，用h数组记录每个集合中的元素，sum记录当加完这条边后此时的u，v点对数目；
程序：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include"stdio.h"
#include"string.h"
#include"iostream"
#include"map"
#include"string"
#include"queue"
#include"stdlib.h"
#include"math.h"
#include"algorithm"
#include"vector"
#define M 100009
#define eps 1e-10
#define inf 1000000000
#define mod 1000000000
#define INF 1000000000
using namespace std;
struct node
{
int u,v,w;
}e[M*5],q[M];
int f[M],h[M];
__int64 sum,ans[M];
int cmp(node a,node b)
{
return a.wy)
{
f[x]=y;
int tt=h[y];
h[y]+=h[x];
sum+=h[y]*(h[y]-1)/2-h[x]*(h[x]-1)/2-tt*(tt-1)/2;
}
else
{
f[y]=x;
int tt=h[x];
h[x]+=h[y];
sum+=h[x]*(h[x]-1)/2-h[y]*(h[y]-1)/2-tt*(tt-1)/2;
}
}
int main()
{
int n,m,k,i;
while(scanf("%d%d%d",&n,&m,&k)!=-1)
{
for(i=0;i