杭电多校第四场-H- K-th Closest Distance

题目描述

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.

输入

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).

输出

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

样例输入

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

样例输出

0
1

题意
给一个序列A，每次询问L，R，p，k，输出[L,R]区间中第k大的|p-a[i]| 

思路
二分答案，查询[L,R]中[p-mid,p+mid]的数的个数
可以用主席树/归并树维护区间x，y之间数的个数（题解说还可以用线段树）
主席树O(mlog1e6logn) 归并树O(mlog1e6logn^2)  
归并树：https://www.cnblogs.com/bennettz/p/8342242.html

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N=1e5+10;
int T,n,m;
int a[N],t[20][N];
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;
}
void build(int s,int l,int r)
{
    if (l==r)
    {
        t[s][l]=a[l];
        return;
    }
    int mid=(l+r)>>1;
    build(s+1,l,mid); build(s+1,mid+1,r);
    for (int i=l,j=mid+1,k=l;i<=mid||j<=r;)
    {
        if (j>r) t[s][k++]=t[s+1][i++];
        else if (i>mid || t[s+1][i]>t[s+1][j]) t[s][k++]=t[s+1][j++];
        else t[s][k++]=t[s+1][i++];
    }
}
int query(int s,int l,int r,int L,int R,int x,int y)
{
    if (x>y ) return 0;
    if (L<=l&r<=R)
    {
        //printf("x=%d,y=%d,l=%d,r=%d,>y=%d,>=x=%d\n",x,y,l,r,upper_bound(t[s]+l,t[s]+r+1,y),upper_bound(t[s]+l,t[s]+r+1,x));
        return upper_bound(t[s]+l,t[s]+r+1,y)-lower_bound(t[s]+l,t[s]+r+1,x);
    }
    int mid=(l+r)>>1,ans=0;
    if (L<=mid) ans+=query(s+1,l,mid,L,R,x,y);
    if (R>mid) ans+=query(s+1,mid+1,r,L,R,x,y);
    return ans;
}
int main()
{
   // freopen("14162.in","r",stdin);
   // freopen("1.out","w",stdout);
    T=read();
    while(T--)
    {
        n=read(); m=read();
        //memset(t,0,sizeof(t));

        for (int i=1;i<=n;i++) a[i]=read();
        build(0,1,n);

        int ans=0;
        int L,R,p,k;
        while (m--)
        {
            L=read(); R=read(); p=read(); k=read();
            L^=ans; R^=ans; p^=ans; k^=ans;
            int l=0,r=1000005;
            while (l<=r)
            {
                int mid=(l+r)>>1;
                //cout<<mid<<endl;
                if (query(0,1,n,L,R,p-mid,p+mid)>=k) ans=mid,r=mid-1;
                else l=mid+1;
            }
            printf("%d\n",ans);
        }
    }
   // fclose(stdin);
   // fclose(stdout);
    return 0;
}

#include <bits/stdc++.h>
using namespace std;
const int N=1e5+10;
int ls[N*21],rs[N*21],s[N*21],root[N];
int a[N],b[N];
int T,n,m,sz;
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;
}
void Insert(int l,int r,int x,int &y,int v)
{
    y=++sz;
    s[y]=s[x]+1;
    if (l==r) return;
    ls[y]=ls[x]; rs[y]=rs[x];
    int mid=(l+r)>>1;
    if (v<=mid) Insert(l,mid,ls[x],ls[y],v);
    else Insert(mid+1,r,rs[x],rs[y],v);
}
int query(int l,int r,int L,int R,int x,int y)
{
    if (L<=l&&r<=R) return s[y]-s[x];
    int ret=0;
    int mid=(l+r)>>1;
    if (L<=mid) ret+=query(l,mid,L,R,ls[x],ls[y]);
    if (R>mid) ret+=query(mid+1,r,L,R,rs[x],rs[y]);
    return ret;
}

int main()
{
    T=read();
    while(T--)
    {
        sz=0;
        n=read(); m=read();
        for (int i=1;i<=n;i++) a[i]=read(),b[i]=a[i];
        sort(b+1,b+1+n);
        int cnt=unique(b+1,b+1+n)-b-1;
        for (int i=1;i<=n;i++)
        {
            a[i]=lower_bound(b+1,b+1+cnt,a[i])-b+1;
            Insert(1,N,root[i-1],root[i],a[i]);
        }
        int ans=0;
        int L,R,p,k;
        while (m--)
        {
            L=read(); R=read(); p=read(); k=read();
            L^=ans,R^=ans,p^=ans,k^=ans;
            int l=0,r=1000006;
            while (l<=r)
            {
                int mid=(l+r)>>1;
                int ll=lower_bound(b+1,b+1+cnt,p-mid)-b+1;
                int rr=upper_bound(b+1,b+1+cnt,p+mid)-b;
                if (query(1,N,ll,rr,root[L-1],root[R])>=k) ans=mid,r=mid-1;
                else l=mid+1;
            }
            printf("%d\n",ans);
        }
    }
    return 0;
}