HDU 6621 (2019杭电第四场 1008) K-th Closest Distance (主席树 + 二分, 求第 k 小绝对值)

#include <bits/stdc++.h>
using namespace std;
const int maxn = 300000;
int sum[maxn*20+5];//sum为森林的不同根结点
struct E {
	int num;
	int l,r;
} t[maxn*20+5]; //t为tree树
int n,sz,m,a[maxn],b[maxn],cnt;
int q;
int build(int l,int r) {
	int now = ++cnt;
	int mid = (l+r)>>1;
	if(l<r) {
		t[now].l = build(l,mid);
		t[now].r = build(mid+1,r);
	}
	return now;
}
int update(int l,int r,int last,int p) {
	int now = ++cnt;
	t[now].num = t[last].num+1;
	t[now].l = t[last].l;
	t[now].r = t[last].r;
	int mid = (l + r)>>1;
	if(l<r) {
		if(p<=mid)t[now].l = update(l,mid,t[last].l,p);
		else t[now].r = update(mid+1,r,t[last].r,p);
	}
	return now;
}
int query(int u,int v,int l,int r,int k){//查询区间第k大 
	if(l==r)return l;
	// v 左孩子包含点个数减去 u 左孩子包含点的个数 
	int tmp = t[t[v].l].num - t[t[u].l].num;
	int mid = (l+r)>>1;
	if(k<=tmp)return query(t[u].l,t[v].l,l,mid,k);
	else return query(t[u].r,t[v].r,mid+1,r,k - tmp);
}
int query2(int u,int v,int l,int r,int k) {//查询区间小于等于k 
	if(k==0)return 0;
	if(l==r)return t[v].num - t[u].num;
	int tmp = t[t[v].l].num - t[t[u].l].num;
	int mid = (l+r)>>1;
	if(k<=mid)return query2(t[u].l,t[v].l,l,mid,k);
	else return tmp+query2(t[u].r,t[v].r,mid+1,r, k);
}
int l,r,pl,pr,p,k,mid,prem,pmid;
void solve(){
	scanf("%d %d %d %d",&pl,&pr,&p,&k);
	pl^=prem;	pr^=prem;	p^=prem;	k^=prem;
	if(pl>pr)swap(pl,pr);
	l = 0;	r = 1e6+5;	pmid = 0;
	while(l<=r){
		mid = (l+r)>>1;
		int t1 = p-mid;
		int t2 = p+mid;
		t1 = lower_bound(b+1,b+1+sz,t1) - b - 1;
		t2 = upper_bound(b+1,b+1+sz,t2) - b - 1;
		if((query2(sum[pl-1],sum[pr],1,sz,t2)-query2(sum[pl-1],sum[pr],1,sz,t1))>=k){
			pmid = mid;
			r = mid-1;
		}else l = mid+1;
	}
	prem = pmid;
	printf("%d\n",prem); 
}
int main() {
	int t;
	scanf("%d",&t);
	
	while(t--) {
		scanf("%d %d",&n,&m);
		for(int i=1; i<=n; i++)scanf("%d",&a[i]),b[i]=a[i];
		sort(b+1,b+n+1);
		sz = unique(b+1,b+n+1) - (b+1);
		cnt = 0;
		sum[0] = build(1,sz);//建立空树
		for(int i=1; i<=n; i++) a[i] = lower_bound(b+1,b+sz+1,a[i]) - b;
		for(int i=1; i<=n; i++) sum[i] = update(1,sz,sum[i-1],a[i]);
		prem = 0;
		while(m--) {
			solve();
		}
	}
	return 0;
}
/*
1
5 2
31 2 5 45 4
2 5 3 2
1 5 5 1

2
6 2
31 31 2 5 45 4
3 5 77 1
1 6 41 2
5 2
31 2 5 45 4
1 3 100 1
2 5 3 2

*/