HDOJ 2665 Kth number
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; }