POJ 2104 K-th Number(划分树)
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K-th Number
Time Limit: 20000MS Memory Limit: 65536KTotal Submissions: 41774 Accepted: 13695Case Time Limit: 2000MS
Description
You are working for Macrohard company in data structures department. After failing your previous task about key insertion you were asked to write a new data structure that would be able to return quickly k-th order statistics in the array segment.
That is, given an array a[1...n] of different integer numbers, your program must answer a series of questions Q(i, j, k) in the form: "What would be the k-th number in a[i...j] segment, if this segment was sorted?"
For example, consider the array a = (1, 5, 2, 6, 3, 7, 4). Let the question be Q(2, 5, 3). The segment a[2...5] is (5, 2, 6, 3). If we sort this segment, we get (2, 3, 5, 6), the third number is 5, and therefore the answer to the question is 5.
That is, given an array a[1...n] of different integer numbers, your program must answer a series of questions Q(i, j, k) in the form: "What would be the k-th number in a[i...j] segment, if this segment was sorted?"
For example, consider the array a = (1, 5, 2, 6, 3, 7, 4). Let the question be Q(2, 5, 3). The segment a[2...5] is (5, 2, 6, 3). If we sort this segment, we get (2, 3, 5, 6), the third number is 5, and therefore the answer to the question is 5.
Input
The first line of the input file contains n --- the size of the array, and m --- the number of questions to answer (1 <= n <= 100 000, 1 <= m <= 5 000).
The second line contains n different integer numbers not exceeding 109 by their absolute values --- the array for which the answers should be given.
The following m lines contain question descriptions, each description consists of three numbers: i, j, and k (1 <= i <= j <= n, 1 <= k <= j - i + 1) and represents the question Q(i, j, k).
The second line contains n different integer numbers not exceeding 109 by their absolute values --- the array for which the answers should be given.
The following m lines contain question descriptions, each description consists of three numbers: i, j, and k (1 <= i <= j <= n, 1 <= k <= j - i + 1) and represents the question Q(i, j, k).
Output
For each question output the answer to it --- the k-th number in sorted a[i...j] segment.
#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; while(~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;}
Sample Input
7 31 5 2 6 3 7 42 5 34 4 11 7 3
Sample Output
563
题意:求区间的第k大数。
划分树模板。
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