静态区间第K小....划分树裸题
Kth number
Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5341 Accepted Submission(s): 1733
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]
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
Source
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=100100;
int tree[20][maxn];
int sorted[maxn];
int toleft[20][maxn];
void build(int l,int r,int dep)
{
if(l==r) return ;
int mid=(l+r)/2;
int same=mid-l+1;
for(int i=l;i<=r;i++)
if(tree[dep][i]<sorted[mid]) same--;
int lpos=l,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)/2;
int cnt=toleft[dep][r]-toleft[dep][l-1];
if(cnt>=k)
{
int newl=L+toleft[dep][l-1]-toleft[dep][L-1];
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()
{
int T_T,n,m;
scanf("%d",&T_T);
while(T_T--)
{
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
{
scanf("%d",sorted+i);
tree[0][i]=sorted[i];
}
sort(sorted+1,sorted+1+n);
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;
}