划分树（求区间第k小值）

区间第K大值查询

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F - Kth number

Time Limit:5000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 2665

Description

Give you a sequence and ask you the kth big number of a inteval.

Input

The first line is the number of the test cases.
For each test case, the first line contain two integer n and m (n, m <= 100000), indicates the number of integers in the sequence and the number of the quaere.
The second line contains n integers, describe the sequence.
Each of following m lines contains three integers s, t, k.
[s, t] indicates the interval and k indicates the kth big number in interval [s, t]

Output

For each test case, output m lines. Each line contains the kth big number.

Sample Input

      
       1 
10 1 
1 4 2 3 5 6 7 8 9 0 
1 3 2

Sample Output

............................................................................此题我有疑问...................................................................

为甚么我输出区间第k小值才能A，而题目问的是第K大的值

划分树模板
求区间的第k大值，或者求给定区间的第k小值
*/

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;

const int MAXN=100010;
int tree[30][MAXN]; //表示划分后每个位置的值，tree[0][1..n]第零层表示输入的原始数据，输入数据不变
int sorted[MAXN]; //sort排序后的输入数据
int toleft[30][MAXN]; //toleft[p][i]表示第i层从1到i有多少个数分入左边

void build(int l,int r,int dep) //建树，build(1, n, 0);
{
if(l==r)return; //到达树的叶子节点
int mid=(l+r)>>1; //中间位置
int same=mid-l+1; //表示等于中间值而且被分入左边的个数***????????????***
for(int i=l;i<=r;i++)
if(tree[dep][i]<sorted[mid])
same--;
int lpos=l; //划分后的左侧开始位置
int rpos=mid+1; //右侧开始位置
for(int i=l;i<=r;i++) //遍历整个数组，进行左右划分
{
if(tree[dep][i]<sorted[mid]) //比中间的数小，分入左边
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 //比中间值大分入右边
tree[dep+1][rpos++]=tree[dep][i];
toleft[dep][i] = toleft[dep][l-1] + lpos - l; //从1到i放左边的个数
}
build(l,mid,dep+1); //对左划分子树递归建树
build(mid+1,r,dep+1); //对优化分子树递归建树
}

/*
*
* 查询区间第k小的数，[L,R]是大区间，[l,r]是要查询的小区间
*
*/
int query(int L,int R,int l,int r,int dep,int k)
{
if(l==r)return tree[dep][l]; //查找到结果
int mid = (L+R)>>1; //划分区间，找中值
int cnt = toleft[dep][r]-toleft[dep][l-1]; //[l,r]中位于左边的个数
if(cnt >= k) //要查询的数在左区间
{
int newl=L+toleft[dep][l-1]-toleft[dep][L-1]; //L+要查询的区间前被放在左边的个数 ????????
int newr=newl+cnt-1; //左端点加上查询区间会被放在左边的个数 ????????
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()
{
freopen("b:\\data.in.txt", "r", stdin);
int T;
int n,m;
int s,t,k;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
memset(tree,0,sizeof(tree));
for(int i=1;i<=n;i++) //从1开始
{
scanf("%d",&tree[0][i]);
sorted[i]=tree[0][i];
}
sort(sorted+1,sorted+n+1);
build(1,n,0); //建树，初始化build(1,n,0)
while(m--)
{
scanf("%d%d%d",&s,&t,&k);
printf("%d\n",query(1,n,s,t,0,k));
}
// printf("排序的数组为：\n");
// for(int i = 1; i <= 10; i++)
// cout << sorted[i] << " ";
// cout << endl << endl;
// for(int i = 0; i < 6; i++)
// {
// printf("第%d层：\n", i);
// for(int j = 1; j <= 10; j++)
// cout << tree[i][j] << " ";
// cout << endl << endl;
// }
}
return 0;
}