POJ 2104 K-th Number（划分树）

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 10 ⁹ 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.

 

题目大意：给一串数字，多次询问区间的第k小值

思路：划分树模板题。关于划分树：http://www.cppblog.com/MatoNo1/archive/2011/06/27/149604.aspx

 

 1 #include <cstdio>

 2 #include <algorithm>

 3 using namespace std;

 4 

 5 #define MAXN 100005

 6 int a[MAXN], a_sort[MAXN];

 7 int n, m;

 8 int sum[20][MAXN];

 9 int tree[20][MAXN];

10 

11 void build(int l, int r, int dep) {

12     int i, mid = (l + r) >> 1, sum_mid = mid - l + 1, lp = l, rp = mid + 1;

13     for(int i = mid - 1; i >= l; --i)

14         if(a_sort[i] < a_sort[mid]) {sum_mid = mid - i; break;}

15     for(i = l; i <= r; ++i) {

16         sum[dep][i] = (i == l ? 0 : sum[dep][i - 1]);

17         if (tree[dep][i] < a_sort[mid]){

18             ++sum[dep][i];

19             tree[dep + 1][lp++] = tree[dep][i];

20         } else if(tree[dep][i] == a_sort[mid] && sum_mid) {

21             --sum_mid;

22             ++sum[dep][i];

23             tree[dep + 1][lp++] = tree[dep][i];

24         } else tree[dep + 1][rp++] = tree[dep][i];

25     }

26     if (l == r)return;

27     build(l, mid, dep + 1);

28     build(mid + 1, r, dep + 1);

29 }

30 

31 int query(int l, int r, int s, int t, int k, int dep) {

32     if(l == r) return tree[dep][l];

33     int mid = (l + r) >> 1;

34     int sum1 = (l == s ? 0 : sum[dep][s - 1]), sum2 = sum[dep][t] - sum1;

35     if(k <= sum2) return query(l, mid, l + sum1, l + sum1 + sum2 - 1, k, dep + 1);

36     else return query(mid + 1, r, mid + s - l + 1 - sum1, mid + t - l + 1 - sum1 - sum2, k - sum2, dep + 1);

37 }

38 

39 int main(){

40     int s, t, k;

41     while(scanf("%d%d", &n, &m) != EOF){

42         for(int i = 1; i <= n; ++i){

43             scanf("%d", &a[i]);

44             tree[0][i] = a_sort[i] = a[i];

45         }

46         sort(a_sort + 1, a_sort + 1 + n);

47         build(1, n, 0);

48         while(m--){

49             scanf("%d%d%d", &s, &t, &k);

50             printf("%d\n", query(1, n, s, t, k, 0));

51         }

52     }

53     return 0;

54 }

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