hdu 2665 划分树

Kth number

Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 7616 Accepted Submission(s): 2402



Problem 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

2

题意：

给你一串数字，然后是m次查询，每次查找[s,t]区间的第k大值 /*划分树学习

/*
模板题，hdu2665
*/

#include <functional>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <Map>
using namespace std;
typedef long long ll;
typedef long double ld;

using namespace std;

const int maxn = 100010;

int tree[20][maxn];
int sorted[maxn];
int toleft[20][maxn];

void build(int l,int r,int dep)  //模拟快排 并记录左树中比i小的个数
{
if(l == r)
return;
int mid = (l+r)>>1;
int same = mid-l+1;//可能有很多值与中间那个相等，但不一定被分到左边
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];  //所查找区间放在左树中的个数
if(cnt >= k)
{   //得到l左边放到左子树的个数，加上L即是开始位置
int lpos = L+toleft[dep][l-1]-toleft[dep][L-1];
int rpos = lpos+cnt-1;
return query(L,mid,lpos,rpos,dep+1,k);
}
else
{   //R-r可以得出后面空出了多少位置
int rpos = r+toleft[dep][R]-toleft[dep][r];
int lpos = rpos-(r-l-cnt);
return query(mid+1,R,lpos,rpos,dep+1,k-cnt);
}
}

int main()
{
int T,n,m;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
memset(tree,0,sizeof(tree));
for(int i = 1;i <= n;i++)
{
scanf("%d",&sorted[i]);
tree[0][i] = sorted[i];
}
sort(sorted+1,sorted+n+1);
build(1,n,0);
int l,r,k;
while(m--)
{
scanf("%d%d%d",&l,&r,&k);
printf("%d\n",query(1,n,l,r,0,k));
}
}
return 0;
}