BZOJ 3545: [ONTAK2010]Peaks [Splay启发式合并]

3545: [ONTAK2010]Peaks

题意:带权图，多组询问与一个点通过边权\(\le x\)的边连通的点中点权k大值

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又读错题了，输出点一直WA，问的是点权啊

本题加强版强制在线了，那这道题肯定离线啊，边权从小到大加边不就是煞笔提吗

奇怪的是合并的时候先序遍历才行...中序和后序都T了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
#define lc t[x].ch[0]
#define rc t[x].ch[1]
#define pa t[x].fa
const int N=1e5+5, M=5e5+5;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}

int n, m, Q, val[N], u, v, w, ans[M];
struct meow{
	int u,v,w;
	bool operator <(const meow &r)const{return w<r.w;}
}a[M];
struct qmeow{
	int u,w,k,id;
	bool operator <(const qmeow &r)const{return w<r.w;}
}q[M];
int fa[N];
int find(int x) {return x==fa[x]?x:fa[x]=find(fa[x]);}

struct SplayTree{
	struct meow{int ch[2], fa, v, size, w;}t[N];
	int sz, root;
	inline void ini() {
		for(int i=1; i<=n; i++) t[i].size=t[i].w=1, t[i].v=val[i];
	}
	inline int wh(int x) {return t[pa].ch[1]==x;}
	inline void update(int x) {t[x].size=t[lc].size+t[rc].size+t[x].w;}
	inline void rotate(int x) {
		int f=t[x].fa, g=t[f].fa, c=wh(x);
		if(g) t[g].ch[wh(f)]=x; t[x].fa=g;
		t[f].ch[c]=t[x].ch[c^1]; t[t[f].ch[c]].fa=f;
		t[x].ch[c^1]=f; t[f].fa=x;
		update(f); update(x);
	}
	inline void splay(int x, int tar) {
		for(; pa!=tar; rotate(x))
			if(t[pa].fa != tar) rotate(wh(x)==wh(pa) ? pa : x);
		if(tar==0) root=x;
	}
	void insert(int p) { 
		int x=root, f=0, v=t[p].v; //printf("insert %d  root %d\n",p,x);
		while(x) {
			if(t[x].v == v) {t[x].w++; t[x].size++; splay(x, 0); return;}
			f=x;
			x = v<t[x].v ? lc : rc;
		}
		x=p;
		t[f].ch[ v>t[f].v ]=x; t[x].fa=f; //printf("f %d %d\n",f,x);
		splay(x, 0);  //see(x);
	}
	void order(int x) {
		int l=lc, r=rc; 
		lc=rc=pa=0; t[x].size=t[x].w;
		if(l) order(l);
		if(r) order(r);
		insert(x);
	}
	void Merge(int x, int y) {
		if(x==y) return; //printf("Merge %d %d\n",x,y);
		splay(x, 0); splay(y, 0);
		if(t[x].size > t[y].size) swap(x, y);
		fa[x]=y; root=y;
		order(x);
	}
	int kth(int x, int k) {
		splay(x, 0); int lsize=0;  //printf("kth %d %d  %d\n",x,k,t[x].size);
		if(k>t[x].size) return -1;
		k=t[x].size-k+1; 
		while(x) {
			int _=lsize+t[lc].size;
			if(k<=_) x=lc;
			else if(k<=_+t[x].w) return t[x].v;
			else lsize=_+t[x].w, x=rc;
		}
		return -1;
	}
}S;

int main() {
	freopen("in","r",stdin);
	n=read(); m=read(); Q=read();
	for(int i=1; i<=n; i++) val[i]=read(), fa[i]=i;
	S.ini();
	for(int i=1; i<=m; i++) u=read(),v=read(),w=read(),a[i]=(meow){u,v,w};
	for(int i=1; i<=Q; i++) u=read(),v=read(),w=read(),q[i]=(qmeow){u,v,w,i};

	sort(a+1,a+1+m); sort(q+1, q+1+Q);
	int now=1;
	for(int i=1; i<=Q; i++) {
		int w=q[i].w;
		while(now<=m && a[now].w<=w) S.Merge(find(a[now].u), find(a[now].v)), now++;
		ans[q[i].id] = S.kth(find(q[i].u), q[i].k);
	}
	for(int i=1; i<=Q; i++) printf("%d\n",ans[i]);
}