主席树K-th Number

最新推荐文章于 2021-06-15 16:18:10 发布

转载最新推荐文章于 2021-06-15 16:18:10 发布 · 89 阅读

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原文链接：http://www.cnblogs.com/xydddd/p/5144142.html

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。文章提供了一个使用C++实现的具体示例，包括数据结构的设计与构建过程。

/*K-th Number
Time Limit: 20000MS Memory Limit: 65536K
Total Submissions: 44535 Accepted: 14779
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*/
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
int ls[8000008],rs[8000008],sum[8000008],root[100008],num[100009],n,m,hash[100009];
int tmp,size,a[100008];
void jia(int l,int r,int x,int &y,int v)
{
y=++size;
sum[y]=sum[x]+1;
if(l==r)
return;
ls[y]=ls[x];
rs[y]=rs[x];
int mid=(l+r)>>1;
if(v<=mid)
jia(l,mid,ls[x],ls[y],v);
else
jia(mid+1,r,rs[x],rs[y],v);
return;
}
int xun(int L,int R,int V)
{
int x=root[L-1],y=root[R],l=1,r=tmp,mid=(l+r)/2;
for(;l!=r;)
if(sum[ls[y]]-sum[ls[x]]>=V)
{
x=ls[x];
y=ls[y];
r=mid;
mid=(l+r)>>1;
}
else
{
V-=sum[ls[y]]-sum[ls[x]];
x=rs[x];
y=rs[y];
l=mid+1;
mid=(l+r)>>1;
}
return l;
}
int main()
{
scanf("%d%d",&n,&m);
for(int i=0;i<n; i++)
{
scanf("%d",&num[i]);
a[i+1]=num[i];
}
sort(num,num+n);
hash[++tmp]=num[0];
for(int i=1;i<n;i++)
if(hash[tmp]!=num[i])
hash[++tmp]=num[i];
for(int i=1;i<=n;i++)
jia(1,tmp,root[i-1],root[i],lower_bound(hash+1,hash+n+1,a[i])-hash);
for(int i=0;i<m;i++)
{
int l,r,v;
scanf("%d%d%d",&l,&r,&v);
printf("%d\n",hash[xun(l,r,v)]);
}
return 0;
}