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=100005,M=500005,QM=2000005;
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=0;char ch=getchar();
    while(ch>'9' || ch<'0')ch=getchar();
    while(ch>='0' && ch<='9')str=(str<<1)+(str<<3)+ch-48,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=1;i<=m;i++){
        e[i].x=gi();e[i].y=gi();e[i].dis=gi();
    }
    for(int i=1;i<=Q;i++)q[i].id=i,q[i].x=gi();
    for(int i=1;i<=n;i++)fa[i]=i,size[i]=1;
    sort(e+1,e+m+1);
    sort(q+1,q+Q+1,compone);
    int cnt=0,sum=0,p=1;
    for(int i=1;i<=Q;i++){
        while(e[p].dis<=q[i].x && cnt<n-1 && 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+1,q+Q+1,comptwo);
    for(int i=1;i<=Q;i++)
        printf("%lld
",q[i].sum);
}
int main()
{
    work();
    return 0;
}