hdu6621 二分加主席树

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6621

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:
冘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

0

1

 

求区间内| a[i]-p | 第k小的数，二分答案，查询区间[ p-ans,p+ans ]内数的数量是否大于等于k

#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
#define Max 1000000
#define maxn 100005
int a[maxn],b[maxn],T[maxn<<],L[maxn<<],R[maxn<<],sum[maxn<<],tot;
inline int update(int pre,int l,int r,int x)
{
    int rt=++tot;
    L[rt]=L[pre];
    R[rt]=R[pre];
    sum[rt]=sum[pre]+;
    if(l<r)
    {
        int mid=l+r>>;
        if(x<=mid)L[rt]=update(L[pre],l,mid,x);
        else R[rt]=update(R[pre],mid+,r,x);
    }
    return rt;
}
inline int query(int u,int v,int ql,int qr,int l,int r)
{
    if(ql<=l&&qr>=r)return sum[v]-sum[u];
    int mid=l+r>>,ans=;
    if(ql<=mid)ans+=query(L[u],L[v],ql,qr,l,mid);
    if(qr>mid)ans+=query(R[u],R[v],ql,qr,mid+,r);
    return ans;
}
int main()
{
    int t;
    cin>>t;
    for(int W=;W<=t;W++)
    {
        int n,m;
        cin>>n>>m;
        for(int i=;i<=n;i++)
        {
            cin>>a[i];
        }
        T[]=sum[]=L[]=R[]=tot=;
        for(int i=;i<=n;i++)
        {
            T[i]=update(T[i-],,Max,a[i]);
        }
        int l,r,p,k,x=;
        for(int i=;i<=m;i++)
        {
            cin>>l>>r>>p>>k;
            l^=x;r^=x;p^=x;k^=x;
            int pl=,pr=Max;
            while(pl<pr)
            {
                int mid=pl+pr>>;
                if(query(T[l-],T[r],max(,p-mid),min(Max,p+mid),,Max)>=k)
                {
                    x=mid;
                    pr=mid;
                }
                else pl=mid+;
            }
            cout<<x<<endl;
        }
    }
    return ;
}