hdu 3938 Portal

对于新增加的每一条边，ab,假如a点和b点原来不连通，那么新增加路径的条数（包含ab的路径）=原来和a点连通的点数（包括自身）*原来和b点连通的点数（包括自身）。

假如a和b已经连通，那么不增加。（为什么不知道，凑数据,我当时题目一直觉得没看懂，因为题目说的非常含糊）

这份题解：写的很清楚

http://blog.csdn.net/sdj222555/article/details/7439187

按照边权值从小到大的顺序给编排序，记录前i条边可以组成的路径数量。

查询时我使用了二分。

Portal

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1250    Accepted Submission(s): 616

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

 

Source

2011 Multi-University Training Contest 10 - Host by HRBEU

 

/**==========================================
 *   This is a solution for ACM/ICPC problem
 *
 *   @source：hdu 3938
 *   @type:  并查集
 *   @author: wust_ysk
 *   @blog:  http://blog.csdn.net/yskyskyer123
 *   @email: [email protected]
 *===========================================*/
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
const int INF =0x3f3f3f3f;
const int maxn= 10000   ;
const int maxm= 50000   ;
int n,m,q;
int pre[maxn+5];
int a[maxn+5];
struct  Edge
{
    int x,y,w;
    Edge(){}
    Edge(int x,int y,int w):x(x),y(y),w(w){}
    bool operator<(const Edge t)const
    {
        return w>1;
        if(edges[mid].w<=x)  le=mid+1;
        else   ri=mid-1;
    }
    return ri;
}
int main()
{
    while(~scanf("%d%d%d",&n,&m,&q))
    {
        int x,y,w;
        init();
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%d",&x,&y,&w);
            edges[i]=Edge(x,y,w);
        }
        sort(edges+1,edges+1+m);
        work();
        for(int i=1;i<=q;i++)
        {
            scanf("%d",&x);
            int p=bs(1,m,x);
//            cout<