离线+并查集 Portal

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。

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：

求有多少条路径使得路径上的花费小于L，这里路径上的花费是这样规定的，a、b两点之间的多条路径中的最长的边最小值！最小生成树--如果两个并查集没有连通，那么联通之后的路径条数就应该是（num[a]*num[b]）........

#include
#include
#include
#include
using namespace std;
int father[10005],num[10005];
struct node
{
    int v1,v2;
    int dis;
}s[50005];
struct node1
{
    int ans;
    int sum;
    int id;
}t[10005];
int cmp(const node a,const node b)
{
    if(a.dis0)
    {
        for(int i=0;i<=n;i++)
        {
            father[i]=i;
            num[i]=1;
        }
        int maxn=0;
        for(int i=0;i=s[j].dis)
            {
                int tmp=s[j].v1;
                int tmp1=s[j].v2;
                tmp=find(tmp);
                tmp1=find(tmp1);
                if(tmp!=tmp1)
                {
                    t[i].ans+=liantong(tmp,tmp1);
                }
                j++;
            }
        }
        sort(t,t+q,cmp2);
        for(int i=0;i