HDU 2665 Kth number（划分树入门题，纯套模板）

Kth number

Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2026 Accepted Submission(s): 672


[align=left]Problem Description[/align]
Give you a sequence and ask you the kth big number of a inteval.

[align=left]Input[/align]
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]

[align=left]Output[/align]
For each test case, output m lines. Each line contains the kth big number.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

2

[align=left]Source[/align]
HDU男生专场公开赛——赶在女生之前先过节（From WHU）

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划分树。模板》。。

/*
HDU  2665 Kth number
划分树

*/

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

const int MAXN=100010;
int tree[30][MAXN];//表示每层每个位置的值
int sorted[MAXN];//已经排序的数
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)
{
if(l==r)return;
int mid=(l+r)>>1;
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;
int rpos=mid+1;
for(int i=l;i<=r;i++)
{
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);
build(mid+1,r,dep+1);

}

//查询区间第k大的数，[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];
int mid=(L+R)>>1;
int cnt=toleft[dep][r]-toleft[dep][l-1];//[l,r]中位于左边的个数
if(cnt>=k)
{
//L+要查询的区间前被放在左边的个数
int newl=L+toleft[dep][l-1]-toleft[dep][L-1];
//左端点加上查询区间会被放在左边的个数
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()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int T;
int n,m;
int s,t,k;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
memset(tree,0,sizeof(tree));//这个必须
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);
while(m--)
{
scanf("%d%d%d",&s,&t,&k);
printf("%d\n",query(1,n,s,t,0,k));
}
}
return 0;
}