poj 2104 -- kth number 平方分割法例题
2015-02-09 11:03
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K-th Number
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.
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).
Output
For each question output the answer to it --- the k-th number in sorted a[i...j] segment.
Sample Input
Sample Output
Hint
This problem has huge input,so please use c-style input(scanf,printf),or you may got time limit exceed.
Source
Northeastern Europe 2004, Northern Subregion
挑战程序设计竞赛上平方分割法的例题。
找出指定区间[i,j]第k小的数。
区间内第k小的数x,即<=x的有>=k个,找到最小的且满足这个条件的x即是答案。二分枚举x得到答案。
用平方分割的做法(线段树当然也可以),将整个区间分成大小为b的一个个桶。每次查找<=x的个数的复杂度为:
1.对于完全包含在区间内的桶,直接二分O(logb),所以事先每个桶要先排序。
2.对于部分包含在区间内的桶,则直接线性查找O(b)
满足1的最多有n/b个桶,满足2的最多只有两个桶,所以每次查找<=x个数的复杂度为O((n/b)*logb+b),要使复杂度取最小,b大约取sqrt(nlogn),复杂度为O(sqrt(nlogn))。再加上二分枚举答案,就是O(sqrt(nlogn)*logn)
同时事先排序预处理复杂度为 O(nlogn)
总体复杂度为O(nlogn+m*sqrt(nlogn)*logn)
Time Limit: 20000MS | Memory Limit: 65536K | |
Total Submissions: 39585 | Accepted: 12912 | |
Case Time Limit: 2000MS |
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.
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).
Output
For each question output the answer to it --- the k-th number in sorted a[i...j] segment.
Sample Input
7 3 1 5 2 6 3 7 4 2 5 3 4 4 1 1 7 3
Sample Output
5 6 3
Hint
This problem has huge input,so please use c-style input(scanf,printf),or you may got time limit exceed.
Source
Northeastern Europe 2004, Northern Subregion
挑战程序设计竞赛上平方分割法的例题。
找出指定区间[i,j]第k小的数。
区间内第k小的数x,即<=x的有>=k个,找到最小的且满足这个条件的x即是答案。二分枚举x得到答案。
用平方分割的做法(线段树当然也可以),将整个区间分成大小为b的一个个桶。每次查找<=x的个数的复杂度为:
1.对于完全包含在区间内的桶,直接二分O(logb),所以事先每个桶要先排序。
2.对于部分包含在区间内的桶,则直接线性查找O(b)
满足1的最多有n/b个桶,满足2的最多只有两个桶,所以每次查找<=x个数的复杂度为O((n/b)*logb+b),要使复杂度取最小,b大约取sqrt(nlogn),复杂度为O(sqrt(nlogn))。再加上二分枚举答案,就是O(sqrt(nlogn)*logn)
同时事先排序预处理复杂度为 O(nlogn)
总体复杂度为O(nlogn+m*sqrt(nlogn)*logn)
#include <iostream> #include <cstring> #include <cstdio> #include <cmath> #include <algorithm> using namespace std; #define maxn 100005 int n, m; int a[maxn], a1[maxn], a2[maxn]; int main() { while(scanf("%d%d", &n, &m)!=EOF){ for(int i = 1; i <= n; i++) scanf("%d", a+i); int b = sqrt(n*log(n+0.0)+0.0); if(!b) b++; memcpy(a1, a, sizeof(a)); memcpy(a2, a, sizeof(a)); for(int i = 1; i+b <= n+1; i+=b) sort(a2+i, a2+i+b); sort(a1+1, a1+n+1); int l, r ,k; for(int t = 0; t < m; t++){ scanf("%d%d%d", &l, &r, &k); int s = 1+(l+b-2)/b*b, e = r/b*b; int ans; int lb=1, ub=n+1, mb; while(lb < ub){ mb = lb+(ub-lb)/2; mb = a1[mb]; int cnt = 0; for(int i = l; i < s && i <= r; i++) if(a[i] <= mb) cnt++; for(int i = e+1; i >= l && i <= r; i++) if(a[i] <= mb) cnt++; for(int i = s; i <= e; i+=b) cnt += (upper_bound(a2+i, a2+i+b, mb)-a2)-i; if(cnt >= k ){ ans = mb; ub = lb+(ub-lb)/2; } else lb = lb+(ub-lb)/2+1; } printf("%d\n", ans); } } return 0; }
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