HDOJ-----3938---Portal并查集
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
带权并查集的题，主要是题意比较费解，搞了半天才弄明白
首先不能一次一次的查找，1e8次查找肯定会爆掉的，所有数据输入完之后，查一次就好
把输入的边按照权值升序排列，要查询的花费也升序排列
因为花费等于一条路的最长边的值，所以按照边的权值升序排列，依次加入图中，即可求得不大于此花费的路径数目
要进行并查集运算的两点可以看做两幅图，两幅图合并所增加的路径数目为两图的节点数a*b
所以用一个数组存节点数，再用一个变量存不大于当前花费的路径数目（累加即可）
另：成环的时候不增加路径，很容易想到，边是递增序，新加入的肯定大于之前的，而题目又要求最长边最小，所以不会经过这条边
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