3545: [ONTAK2010]Peaks 平衡树，最小生成树

链接

https://www.lydsy.com/JudgeOnline/problem.php?id=3545

离线询问，按照权值排个序
就是在克鲁斯卡尔时候维护个treap，到时候挨个查询一下就好了
nb的gzy说要要在线才是呢，nb

代码

/**************************************************************
    Problem: 3545
    User: gryz2016
    Language: C++
    Result: Accepted
    Time:8708 ms
    Memory:69660 kb
****************************************************************/
 
#include <iostream>
#include <ctime>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll long long
using namespace std;
const int N=1e5+7;
const int inf=0x3f3f3f3f;
int read() {
    int x=0,f=1;char s=getchar();
    for(;s>'9'||s<'0';s=getchar()) if(s=='-') f=-1;
    for(;s>='0'&&s<='9';s=getchar()) x=x*10+s-'0';
    return x*f;
}
int ch[N*40][2],siz[N*40],cnt;
void build(int l,int r,int L,int &now) {
    if(!now) now=++cnt;
    siz[now]++;
    if(l==r) return;
    int mid=(l+r)>>1;
    if(L<=mid) build(l,mid,L,ch[now][0]);
    else build(mid+1,r,L,ch[now][1]);
}
int merge(int x,int y) {
    if(!x||!y) return x+y;
    siz[x]+=siz[y];
    ch[x][0]=merge(ch[x][0],ch[y][0]);
    ch[x][1]=merge(ch[x][1],ch[y][1]);
    return x;
}
int query(int l,int r,int k,int rt) {
    if(l==r) return l;
    int mid=(l+r)>>1;
    if(siz[ch[rt][0]]>=k) return query(l,mid,k,ch[rt][0]);
    else return query(mid+1,r,k-siz[ch[rt][0]],ch[rt][1]);
}
struct node {int u,v,q;}e[N*5];
bool cmp1(node a,node b) {return a.q<b.q;}
struct QQQ {int v,x,k,id;}Q[N*5];
bool cmp2(QQQ a,QQQ b) {return a.x<b.x;}
int n,m,q,a[N*5],f[N*5],ans[N*5],rt[N*5];
int find(int x) {return f[x]==x ? x : f[x]=find(f[x]);}
void uu(int x,int y) {
    if(x==y) return;
    rt[y]=merge(rt[x],rt[y]);
    f[y]=x;
}
int main() {
    // freopen("1.in","r",stdin);
    // freopen("a.out","w",stdout);
    n=read(),m=read(),q=read();
    for(int i=1;i<=n;++i) a[i]=read();
    for(int i=1;i<=m;++i) e[i].u=read(),e[i].v=read(),e[i].q=read();
    for(int i=1;i<=q;++i) Q[i].v=read(),Q[i].x=read(),Q[i].k=read(),Q[i].id=i;
    sort(e+1,e+1+m,cmp1);
    sort(Q+1,Q+1+q,cmp2);
    for(int i=1;i<=n;++i) f[i]=i,build(1,1e9,a[i],rt[i]);
    int js=1;
    memset(ans,-1,sizeof(ans));
    for(int i=1;i<=m;++i) {
        int fx=find(e[i].u),fy=find(e[i].v);    
        if(fx==fy) continue;
        while(Q[js].x<e[i].q&&js<=q) {
            int fa=find(Q[js].v);
            if(siz[rt[fa]]>=Q[js].k)
                ans[Q[js].id]=query(1,1e9,siz[rt[fa]]-Q[js].k+1,rt[fa]);
            js++;
        }
        uu(fx,fy);
    }
    while(Q[js].k<=e[n].q&&js<=q) {
        int fa=find(Q[js].v);
        if(siz[rt[fa]]>=Q[js].k) 
            ans[Q[js].id]=query(1,1e9,siz[rt[fa]]-Q[js].k+1,rt[fa]);
        js++;
    }
    for(int i=1;i<=q;++i) printf("%d\n", ans[i]);
    return 0;
}