POJ 2104 K-th Number(主席树)

最新推荐文章于 2019-08-24 17:42:02 发布

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本文介绍了一种解决K-thNumber问题的方法，该问题要求在给定数组的指定区间内找到排序后的第K个数。通过构建特殊的数据结构，可以高效地解决这一问题。

K-th Number

Time Limit: 20000MS		Memory Limit: 65536K
Total Submissions:66103		Accepted: 23288
Case Time Limit: 2000MS

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 10⁹ 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

Hint

This problem has huge input,so please use c-style input(scanf,printf),or you may got time limit exceed.

#include<cstdio>
#include<algorithm>
#include<string.h>
using namespace std;
const int maxn=100005;
int a[maxn],has[maxn],T[maxn];
int tol,L[maxn*20],R[maxn*20],sum[maxn*20];
int build(int l,int r)
{
    int rt=++tol;
    if(l<r)
    {
        int mid=(l+r)>>1;
        L[rt]=build(l,mid);
        R[rt]=build(mid+1,r);
    }
    return rt;
}
int query(int u,int v,int l,int r,int k)
{
    if(l>=r)return l;
    int s=sum[L[v]]-sum[L[u]];
    int mid=(l+r)>>1;
    if(s>=k) return query(L[u],L[v],l,mid,k);
    else return query(R[u],R[v],mid+1,r,k-s);
}
int update(int pre,int l,int r,int x)
{
    int rt=++tol;
    L[rt]=L[pre];R[rt]=R[pre];sum[rt]=sum[pre]+1;
    if(l<r)
    {
        int mid=(l+r)>>1;
        if(x<=mid) L[rt]=update(L[pre],l,mid,x);
        else R[rt]=update(R[pre],mid+1,r,x);
    }
    return rt;
}
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        has[i]=a[i];
    }
    sort(has+1,has+n+1);
    int d=unique(has+1,has+n+1)-has-1;
    T[0]=build(1,d);
    for(int i=1;i<=n;i++)
    {
        int x=lower_bound(has+1,has+d+1,a[i])-has;
        T[i]=update(T[i-1],1,d,x);
    }
    while(m--)
    {
        int l,r,k;scanf("%d%d%d",&l,&r,&k);
        printf("%d\n",has[query(T[l-1],T[r],1,d,k)]);
    }
    return 0;
}