You are given a weighted tree consisting of n vertices. Recall that a tree is a connected graph without cycles. Vertices ui and vi are connected by an edge with weight wi.
You are given m queries. The i-th query is given as an integer qi. In this query you need to calculate the number of pairs of vertices (u,v) (u<v) such that the maximum weight of an edge on a simple path between u and v doesn't exceed qi.
题意： 给出一个数问都多少个二元组满足 之间的最大边不超过这个数的
解
这个显然满足单调性
所以我们考虑离线 由小到大 加边
答案的贡献为 每个联通 块 size size*(size-1)/2;
并查集维护联通块的点的个数
//
#include<bits/stdc++.h>
using namespace std;
#define maxnn 300000
#define ll long long
ll n,m;
struct node
{
ll st,en;
ll val;
}edge[maxnn];
bool cmp(node a,node b)
{
return a.val<b.val;
}
ll tot=0;
struct gr{
ll id,va;
}gra[maxnn];
bool cmp1(gr a,gr b)
{
return a.va<b.va;
}
ll f[maxnn];
ll ask[maxnn];
ll siz[maxnn];
ll gf(ll v)
{
return f[v]==v?v:f[v]=gf(f[v]);
}
int main()
{
cin>>n>>m;
ll x,y,z;
for(int i=1;i<n;i++)
{
scanf("%lld%lld%lld",&x,&y,&z);
edge[++tot].st=x;
edge[tot].en=y;
edge[tot].val=z;
}
sort(edge+1,edge+1+tot,cmp);
for(int i=1;i<=m;i++)
{
scanf("%lld",&gra[i].va);
gra[i].id=i;
}
sort(gra+1,gra+1+m,cmp1);
ll sec=0;
ll j=1;
ll tmp=0;
for(int i=1;i<=n;i++) f[i]=i,siz[i]=1;
for(int i=1;i<=m;i++)
{
int tr=gra[i].va;
for(;j<=tot;j++)
{
if(edge[j].val<=tr)
{
if(gf(edge[j].st)!=gf(edge[j].en))
{
tmp-=(siz[gf(edge[j].en)]*(siz[gf(edge[j].en)]-1)/2);
tmp-=(siz[gf(edge[j].st)]*(siz[gf(edge[j].st)]-1)/2);
siz[gf(edge[j].en)]+=siz[gf(edge[j].st)];
tmp+=(siz[gf(edge[j].en)]*(siz[gf(edge[j].en)]-1)/2);
f[gf(edge[j].st)]=gf(edge[j].en);
}
}
else break;
}
j--;
ask[gra[i].id]=tmp;
}
for(int i=1;i<=m;i++)
{
cout<<ask[i]<<" ";
}
}