Path Queries （并查集+ 把图/树 分成很多块来看）

You are given a weighted tree consisting of nn vertices. Recall that a tree is a connected graph without cycles. Vertices u_iu 
i
​
  and v_iv 
i
​
  are connected by an edge with weight w_iw 
i
​
 .

You are given mm queries. The ii-th query is given as an integer q_iq 
i
​
 . In this query you need to calculate the number of pairs of vertices (u, v)(u,v) (u < vu<v) such that the maximum weight of an edge on a simple path between uu and vv doesn't exceed q_iq 
i
​
 .

Input
The first line of the input contains two integers nn and mm (1 \le n, m \le 2 \cdot 10^51≤n,m≤2⋅10 
5
 ) — the number of vertices in the tree and the number of queries.

Each of the next n - 1n−1 lines describes an edge of the tree. Edge ii is denoted by three integers u_iu 
i
​
 , v_iv 
i
​
  and w_iw 
i
​
  — the labels of vertices it connects (1 \le u_i, v_i \le n1≤u 
i
​
 ,v 
i
​
 ≤n, u_i \ne v_iu 
i
​
 

=v 
i
​
 ) and the weight of the edge (1 \le w_i \le 2 \cdot 10^51≤w 
i
​
 ≤2⋅10 
5
 ). It is guaranteed that the given edges form a tree.

The last line of the input contains mm integers q_1, q_2, \dots, q_mq 
1
​
 ,q 
2
​
 ,…,q 
m
​
  (1 \le q_i \le 2 \cdot 10^51≤q 
i
​
 ≤2⋅10 
5
 ), where q_iq 
i
​
  is the maximum weight of an edge in the ii-th query.

Output
Print mm integers — the answers to the queries. The ii-th value should be equal to the number of pairs of vertices (u, v)(u,v) (u < vu<v) such that the maximum weight of an edge on a simple path between uu and vv doesn't exceed q_iq 
i
​
 .

Queries are numbered from 11 to mm in the order of the input.

Sample 1
Inputcopy    Outputcopy
7 5
1 2 1
3 2 3
2 4 1
4 5 2
5 7 4
3 6 2
5 2 3 4 1
21 7 15 21 3 
Sample 2
Inputcopy    Outputcopy
1 2
1 2
0 0 
Sample 3
Inputcopy    Outputcopy
3 3
1 2 1
2 3 2
1 3 2

#include <bits/stdc++.h>
using namespace std;
#define ri register int
#define M 200005

template <class G> void read(G &x)
{
    x=0;int f=0;char ch=getchar();
    while(ch<'0'||ch>'9'){f|=ch=='-';ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
    x=f?-x:x;
    return ;
 }

int fa[M];
int n,m;
struct dian{
    int u,v;
    int val;
    bool operator <(const dian &t)const
    {
        return val<t.val;
    }
}p[M];

int find(int a)
{
    if(a==fa[a]) return a;
    fa[a]=find(fa[a]);
    return fa[a];
}
long long ans[M],num[M];
int main(){
    
    read(n);read(m);
    for(ri i=1;i<n;i++)
    {
        int a,b,c;
        read(a);read(b);read(c);
        p[i].u=a;p[i].v=b;p[i].val=c;
    }
    sort(p+1,p+n);
    long long tmp=0;
    for(ri i=1;i<=n;i++) fa[i]=i,num[i]=1;
    for(ri i=1;i<n;i++)
    {
        int l=find(p[i].u);int r=find(p[i].v);
        fa[l]=r;
        long long ff=num[l]*num[r];
        ans[p[i].val]=ff+tmp;
        tmp+=ff;
        num[r]+=num[l];
    }
    for(ri i=1;i<=200000;i++)
    {
        if(ans[i]) continue;
        else ans[i]=ans[i-1];
    }
    for(ri i=1;i<=m;i++)
    {
        int a;
        read(a);
        printf("%lld ",ans[a]);
    }
    return 0;
    
    
    
    
    
}

思路：

• 把树分成很多个点，依据边的大小，一次连接，形成更大的块，连接时维护答案即可（并查集）；