HDU 6621 K-th Closest Distance（2019HDU多校赛第4场）

2019-08-02

阅读量 25

Time Limit: 20000/15000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)

Problem Description

You have an array: a1, a2, …, an and you must answer for some queries.
For each query, you are given an interval [L, R] and two numbers p and K. Your goal is to find the Kth closest distance between p and aL, aL+1, …, aR.
The distance between p and ai is equal to |p - ai|.
For example:
A = {31, 2, 5, 45, 4 } and L = 2, R = 5, p = 3, K = 2.
|p - a2| = 1, |p - a3| = 2, |p - a4| = 42, |p - a5| = 1.
Sorted distance is {1, 1, 2, 42}. Thus, the 2nd closest distance is 1.

Input

The first line of the input contains an integer T (1 <= T <= 3) denoting the number of test cases.
For each test case:
The first line contains two integers n and m (1 <= n, m <= 10^5) denoting the size of array and number of queries.
The second line contains n space-separated integers a1, a2, …, an (1 <= ai <= 10^6). Each value of array is unique.
Each of the next m lines contains four integers L’, R’, p’ and K’.
From these 4 numbers, you must get a real query L, R, p, K like this:
L = L’ xor X, R = R’ xor X, p = p’ xor X, K = K’ xor X, where X is just previous answer and at the beginning, X = 0.
(1 <= L < R <= n, 1 <= p <= 10^6, 1 <= K <= 169, R - L + 1 >= K).

Output

For each query print a single line containing the Kth closest distance between p and aL, aL+1, …, aR.

Sample Input

1
5 2
31 2 5 45 4
1 5 5 1
2 5 3 2

Sample Output

1
2

0
1

思路分析

题目意思是说，给定一段序列，每次求区间[L,R]的数与P的差的绝对值，对其排序后的第K个差值是多少。（后面的L,R,P,K要用之前的答案异或得到，所以一旦前面错了，后面就Game Over了，甚至会RE【别问我为什么quq】）

首先考虑到时间限制，虽然给了15s看上去绰绰有余，实际上偌大的数据量还是不能暴力去算，所以就想到了线段树这个万能的东西！它可以把查询压缩到O(logn)，但这样复杂度还是有点高，所以再考虑二分一下以p为中心的半径，即每次验证[p-mid,p+mid]区间内的数的个数是不是符合要求，这样总体的复杂度就是O(mloglogn)，是不是感觉很赞呢~

用主席树（一说叫权值线段树）查询十分方便，我们只需要以数据初始的id为基建树，然后再一阵基本操作就行！

反正具体的看代码就行！

Code

#include <bits/stdc++.h>
const int maxn=2*1e5+100; 
using namespace std;
int n,m,cnt=0,xk,tt,xl,xr,xp;

int a[maxn],root[maxn],id[maxn];
struct node{
    int l,r,sum;
}T[maxn*40];
void update(int l,int r,int &x,int y,int pos){
    T[++cnt]=T[y];T[cnt].sum++;x=cnt;
    if (l==r) return ;
    int mid=(l+r)>>1;
    if (pos<=mid) update(l,mid,T[x].l,T[y].l,pos);    
    else update(mid+1,r,T[x].r,T[y].r,pos);
}
void ask(int l,int r,int x,int y){
    if (tt>=xk) return;
    if (xl<=l&&r<=xr){
        tt+=T[y].sum-T[x].sum;
        return;
    }
    int mid=(l+r)>>1;
    if (xl<=mid) ask(l,mid,T[x].l,T[y].l);
    if (mid<xr) ask(mid+1,r,T[x].r,T[y].r);
}
bool cmp(int x,int y){
    return a[x]<a[y];
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int test;
    cin>>test;
	while (test--){
    	memset(root,0,sizeof root);
    	memset(T,0,sizeof T);
    	memset(a,0,sizeof a);
    	memset(id,0,sizeof id);
    	cnt=0;
    	cin>>n>>m;
    	for (int i=1;i<=n;i++) {
            cin>>a[i];
            id[i]=i;
   		}
    	sort(id+1,id+1+n,cmp);
    	sort(a+1,a+1+n);
    
    	for (int i=1;i<=n;i++)
            update(1,n,root[i],root[i-1],id[i]);
    	int ans=0;
    	while (m--){
        	cin>>xl>>xr>>xp>>xk;
        	xl^=ans,xr^=ans,xp^=ans,xk^=ans;
        	int ll=-1,rr=1e7;
        	while(rr-ll>1){
            	int mid=(ll+rr)>>1;
            	int ls=lower_bound(a+1,a+n+1,xp-mid)-a;
            	int rs=upper_bound(a+1,a+n+1,xp+mid)-a-1;
            	tt=0;
            	ask(1,n,root[ls-1],root[rs]);
            	if (tt>=xk) rr=mid;else ll=mid;
        	}
        	ans=rr;
        	cout<<ans<<endl; 
    	}
	}
    return 0;    
}