bzoj3545 [ONTAK2010]Peaks（离线+kruskal+主席树+线段树合并）

首先我们可以离线做。把询问按x排序，给每个点建一棵权值线段树。然后一边跑mst，一边维护线段树。每次合并两个点的时候，就要合并两棵线段树。合并的时候要小心因为合并叶子时不能pushup，所以还是每次相加比较稳妥。然后合并的总复杂度应该就是合并时删掉的节点数，是O(nlogn)的，可以接受。所以总复杂度O(nlogn+mlogm)

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define N 100010
#define M 500010
inline int read(){
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
return x*f;
}
int n,nn,m,qq,a
,b
,num=0,h
,ans[M],owo=0,fa
,rt
;
struct Data{
int x,y,val;
}e[M];
inline bool cmp(Data x,Data y){return x.val<y.val;}
struct que{
int v,x,k,id;
}q[M];
inline bool cmp1(que x,que y){return x.x<y.x;}
struct node{
int x,lc,rc;
}tree[N*20];
inline int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}
inline void change(int &p,int l,int r,int x){
tree[++owo]=tree[p];p=owo;tree[p].x++;
if(l==r) return;int mid=l+r>>1;
if(x<=mid) change(tree[p].lc,l,mid,x);
else change(tree[p].rc,mid+1,r,x);
}
inline int merge(int p1,int p2){
if(!p1||!p2) return p1+p2;
tree[p1].lc=merge(tree[p1].lc,tree[p2].lc);
tree[p1].rc=merge(tree[p1].rc,tree[p2].rc);
tree[p1].x+=tree[p2].x;return p1;
}
inline int ask(int p,int l,int r,int k){
if(l==r) return l;int mid=l+r>>1;
if(k<=tree[tree[p].lc].x) return ask(tree[p].lc,l,mid,k);
else return ask(tree[p].rc,mid+1,r,k-tree[tree[p].lc].x);
}
int main(){
//  freopen("a.in","r",stdin);
nn=read();m=read();qq=read();
for(int i=1;i<=nn;++i) a[i]=b[i]=read(),fa[i]=i;
sort(b+1,b+nn+1);n=unique(b+1,b+nn+1)-b-1;
for(int i=1;i<=nn;++i) a[i]=lower_bound(b+1,b+n+1,a[i])-b;
for(int i=1;i<=nn;++i) change(rt[i],1,n,a[i]);
for(int i=1;i<=m;++i) e[i].x=read(),e[i].y=read(),e[i].val=read();
for(int i=1;i<=qq;++i) q[i].v=read(),q[i].x=read(),q[i].k=read(),q[i].id=i;
sort(e+1,e+m+1,cmp);sort(q+1,q+qq+1,cmp1);int now=1;
for(int i=1;i<=qq;++i){
while(now<=m&&e[now].val<=q[i].x){
int x=find(e[now].x),y=find(e[now].y);
if(x!=y){
fa[x]=y;rt[y]=merge(rt[x],rt[y]);
}now++;
}if(tree[rt[find(q[i].v)]].x<q[i].k) ans[q[i].id]=-1;
else ans[q[i].id]=b[ask(rt[fa[q[i].v]],1,n,tree[rt[fa[q[i].v]]].x-q[i].k+1)];
}for(int i=1;i<=qq;++i) printf("%d\n",ans[i]);
return 0;
}