POJ 2832 How Many Pairs?

Description

You are given an undirected graph G with N vertices and M edges. Each edge has a length. Below are two definitions.

1. Define max_len(p) as the length of the edge with the maximum length of p where p is an arbitrary non-empty path in G.
2. Define min_pair(u, v) as min{max_len(p) | p is a path connecting the vertices u and v.}. If there is no paths connecting u and v, min_pair(u, v) is defined as infinity.

Your task is to count the number of (unordered) pairs of vertices u and v satisfying the condition that min_pair(u, v) is not greater than a given integer A.

Input

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 vertices, 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 < 10⁸) describing an edge connecting the vertices a and b with length c. Each of the following Q lines gives a query consisting of a single integer A (0 ≤ A < 10⁸).

Output

Output the answer to each query on a separate line.

Sample Input

4 5 4
1 2 1
2 3 2
2 3 5
3 4 3
4 1 4
0
1
3
2

Sample Output

0
1
6
3

题解：

将边和询问都按从小到大排序，然后对于一组询问，我们枚举所有小于当前询问的边，然后把边的两个端点对应的集合进行计算，并查集合并维护

 #include <algorithm>
 #include <iostream>
 #include <cstdlib>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 using namespace std;
 const int N=,M=,QM=;
 typedef long long ll;
 struct node{
     int x,y,dis;
     bool operator <(const node &pp)const{
         return dis<pp.dis;
     }
 }e[M];
 int gi(){
     int str=;char ch=getchar();
     while(ch>'' || ch<'')ch=getchar();
     while(ch>='' && ch<='')str=(str<<)+(str<<)+ch-,ch=getchar();
     return str;
 }
 int n,m,Q,size[N],fa[N];
 int find(int x){
     return fa[x]==x?x:fa[x]=find(fa[x]);
 }
 struct Question{
     int id,x;ll sum;
 }q[QM];
 bool compone(const Question &pp,const Question &qq){
     return pp.x<qq.x;
 }
 bool comptwo(const Question &pp,const Question &qq){
     return pp.id<qq.id;
 }
 void work(){
     int x,y,dis;
     n=gi();m=gi();Q=gi();
     for(int i=;i<=m;i++){
         e[i].x=gi();e[i].y=gi();e[i].dis=gi();
     }
     for(int i=;i<=Q;i++)q[i].id=i,q[i].x=gi();
     for(int i=;i<=n;i++)fa[i]=i,size[i]=;
     sort(e+,e+m+);
     sort(q+,q+Q+,compone);
     int cnt=,sum=,p=;
     for(int i=;i<=Q;i++){
         while(e[p].dis<=q[i].x && cnt<n- && p<=m){
             x=e[p].x;y=e[p].y;
             if(find(x)==find(y)){
                p++;continue;
             }
             sum+=(ll)size[find(y)]*size[find(x)];
            size[find(x)]+=size[find(y)];
             fa[find(y)]=find(x);
             p++;cnt++;
         }
         q[i].sum=sum;
     }
     sort(q+,q+Q+,comptwo);
     for(int i=;i<=Q;i++)
         printf("%lld\n",q[i].sum);
 }
 int main()
 {
     work();
     return ;
 }