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<