hdu3938（离线并查集）

Portal

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1930    Accepted Submission(s): 961

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：在给定的权值下，满足点对的数目的情况。所谓的要求就是，给出两点，之间有很多路径，这个点对的最小花费就是

众多路径中，最小花费的路径，路径长度就是路径上最长边的长度。

解析：很容易想到离线化输出（离线化不是离散化，离线化是指将所有数据获得，然后一次性输出结果，就像不是在线解决的一样。）认真观察可以看出，两点可以连线的条件就是之前的边都小于给出的定值。于是，利用最小生成树的算法构造边，如果两个集合a，b因为一条边的加入，那么增加的点对数位sum[a]*sum[b](这点自行理解，画图就可以看出来的)，构造边利用到最小

生成树，则需要将对边排序。

#include
using namespace std;

const int maxn=50005;
int n,m,q;
int f[maxn],sum[maxn];
struct node
{
	int v,u,l;
}e[maxn];
struct Q
{
	int l,id,ans;
}que[10005];
bool cmp(node a,node b)  
{  
    return a.l