Description

You are working for Macrohard company in data structures department. After failing your previous task about key insertion you were asked to write a new data structure that would be able to return quickly k-th order statistics in the array segment.

That is, given an array a[1…n] of different integer numbers, your program must answer a series of questions Q(i, j, k) in the form: “What would be the k-th number in a[i…j] segment, if this segment was sorted?”

For example, consider the array a = (1, 5, 2, 6, 3, 7, 4). Let the question be Q(2, 5, 3). The segment a[2…5] is (5, 2, 6, 3). If we sort this segment, we get (2, 3, 5, 6), the third number is 5, and therefore the answer to the question is 5.

Input

The first line of the input file contains n — the size of the array, and m — the number of questions to answer (1 <= n <= 100 000, 1 <= m <= 5 000).

The second line contains n different integer numbers not exceeding 10^9 by their absolute values — the array for which the answers should be given.

The following m lines contain question descriptions, each description consists of three numbers: i, j, and k (1 <= i <= j <= n, 1 <= k <= j - i + 1) and represents the question Q(i, j, k).

Output

For each question output the answer to it — the k-th number in sorted a[i…j] segment.

Sample Input

7 3
1 5 2 6 3 7 4
2 5 3
4 4 1
1 7 3

Sample Output

5
6
3

題意

靜態查詢區間第 k 小的數。

思路

【劃分樹/主席樹】 的基本應用，以下附模板。

AC 程式碼

劃分樹

#include <iostream>
#include<algorithm>
#include<cstring>
using namespace std;
typedef __int64 LL;
const int maxn=1e5+10;

int tree[20][maxn];     // 每層每個位置的值
int sorted[maxn];       // 已經排序好的數
int toleft[20][maxn];   // 第 i 層從 1-j 有多少個數被劃分到了左邊

void build(int l,int r,int dep)
{
    if(l==r)return;
    int mid = (l+r)>>1;
    int same = mid-l+1;     // 等於中間值且被劃分到左邊數的個數
    for(int i=l; i<=r; i++)
    {
        if(tree[dep][i]<sorted[mid])
            same--;
    }
    int lpos = l;
    int rpos = mid+1;
    for(int i=l; i<=r; i++)
    {
        if(tree[dep][i]<sorted[mid])
            tree[dep+1][lpos++]=tree[dep][i];
        else if(tree[dep][i]==sorted[mid]&&same>0)
            tree[dep+1][lpos++]=tree[dep][i],same--;
        else
            tree[dep+1][rpos++]=tree[dep][i];
        toleft[dep][i]=toleft[dep][l-1]+lpos-l;
    }
    build(l,mid,dep+1);
    build(mid+1,r,dep+1);
}

int query(int L,int R,int l,int r,int dep,int k)
{
    if(l==r)return tree[dep][l];
    int mid = (L+R)>>1;
    int cnt = toleft[dep][r]-toleft[dep][l-1];
    if(cnt>=k)
    {
        int newl = L+toleft[dep][l-1]-toleft[dep][L-1];
        int newr = newl+cnt-1;
        return query(L,mid,newl,newr,dep+1,k);
    }
    else
    {
        int newr = r +toleft[dep][R]-toleft[dep][r];
        int newl = newr-(r-l-cnt);
        return query(mid+1,R,newl,newr,dep+1,k-cnt);
    }
}

int main()
{
    ios::sync_with_stdio(false);
    int n,m;
    while(cin>>n>>m)
    {
        memset(tree,0,sizeof(tree));
        for(int i=1; i<=n; i++)
        {
            cin>>tree[0][i];
            sorted[i]=tree[0][i];
        }
        sort(sorted+1,sorted+n+1);
        build(1,n,0);
        int s,t,k;
        while(m--)
        {
            cin>>s>>t>>k;
            cout<<query(1,n,s,t,0,k)<<endl;
        }
    }
    return 0;
}

主席樹

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef __int64 LL;
const int maxn = 2000000;

int a[maxn],b[maxn];
int sum[maxn];
int ls[maxn];
int rs[maxn];
int rk[maxn];
int tot;

void build(int &o,int l,int r)
{
    o=++tot;
    sum[o]=0;
    if(l==r)return;
    int mid = (l+r)>>1;
    build(ls[o],l,mid);
    build(rs[o],mid+1,r);
}

void update(int &o,int l,int r,int last,int p)
{
    o = ++tot;
    ls[o]=ls[last];
    rs[o]=rs[last];
    sum[o]=sum[last]+1;
    if(l==r)return;
    int mid = (l+r)>>1;
    if(p<=mid)update(ls[o],l,mid,ls[last],p);
    else update(rs[o],mid+1,r,rs[last],p);
}

int query(int L,int R,int l,int r,int k)
{
    if(L==R)return L;
    int mid = (L+R)>>1;
    int cnt = sum[ls[r]]-sum[ls[l]];
    if(k<=cnt)return query(L,mid,ls[l],ls[r],k);
    else return query(mid+1,R,rs[l],rs[r],k-cnt);
}

int main()
{
    ios::sync_with_stdio(false);
    int n,m;
    while(cin>>n>>m)
    {
        tot=0;
        for(int i=1; i<=n; i++)
            cin>>a[i],b[i]=a[i];
        sort(b+1,b+n+1);
        int sz = unique(b+1,b+n+1)-(b+1);
        build(rk[0],1,sz);
        for(int i=1; i<=n; i++)
        {
            a[i] = lower_bound(b+1,b+n+1,a[i])-b;
            update(rk[i],1,sz,rk[i-1],a[i]);
        }
        while(m--)
        {
            int l,r,k;
            cin>>l>>r>>k;
            cout<<b[query(1,sz,rk[l-1],rk[r],k)]<<endl;
        }
    }
    return 0;
}