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链接：
题目：
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.
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 the answer to each query on a separate line.
10 10 107 2 16 8 34 5 85 8 22 8 96 4 52 1 58 10 57 3 77 8 810615918276
分析与总结：
做这题学到了什么是“离线算法”的概念。所谓“离线”，就是把所有的数据都输入之后再计算，“在线”就是边输入边计算。
用在这题中，是因为输入中的“询问部分”，有Q 个问，每个L可以有多少种不同路径。由于大的L必定会包含到小的L， 所以把所有问题都输入，再从大到小排序，再计算，可以减少很多计算量。
这题还需要用到的是并查集中的“权值”， 用rank数组表示，也就是某个棵树k有rank【k】个结点。同一个树之间的点都是连通的，任何点都可以通往其它的任意点， 那么当两颗树合并成一棵树时， 将会增加rank[a]*rank[b]条路径。
代码：
#include #include using namespace std;#define N 10005int f[N], rank[N], ans[N], n, m, Q;struct Edge{ int u, v, val; friend bool operator < (const Edge&a,const Edge&b){ return a.val < b.val; }}arr[N*5];struct Query{ int id, L; friend bool operator<(const Query&a,const Query&b){ return a.L
—— 生命的意义，在于赋予它意义。 原创 ， By D_Double (转载请标明)