hdu 2665 Kth number（划分树）

Kth number

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6342    Accepted Submission(s): 2021

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

 

Source

HDU男生专场公开赛——赶在女生之前先过节（From WHU）

 

Recommend

zty

 

求区间第k大。

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<cmath>#define N 5050using namespace std;const int MAXN = 100010;int tree[20][MAXN];//表示每层每个位置的值int sorted[MAXN];//已经排序好的数int toleft[20][MAXN];//toleft[p][i]表示第i层从1到i有数分入左边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++) //注意是l,不是one        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);}//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];    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() {   // freopen("test.in","r",stdin);    int n,m;    int t;    cin>>t;    while(t--) {        scanf("%d%d",&n,&m);        memset(tree,0,sizeof(tree));        for(int i = 1; i <= n; i++) {            scanf("%d",&tree[0][i]);            sorted[i] = tree[0][i];        }        sort(sorted+1,sorted+n+1);        build(1,n,0);        int s,t,k;        while(m--) {            scanf("%d%d%d",&s,&t,&k);            printf("%d\n",query(1,n,s,t,0,k));        }    }    return 0;}