HDU 4251 The Famous ICPC Team Again

The Famous ICPC Team Again

Time Limit: 30000/15000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Problem Description

When Mr. B, Mr. G and Mr. M were preparing for the 2012 ACM-ICPC World Final Contest, Mr. B had collected a large set of contest problems for their daily training. When they decided to take training, Mr. B would choose one of them from the problem set. All the problems in the problem set had been sorted by their time of publish. Each time Prof. S, their coach, would tell them to choose one problem published within a particular time interval. That is to say, if problems had been sorted in a line, each time they would choose one of them from a specified segment of the line.

Moreover, when collecting the problems, Mr. B had also known an estimation of each problem’s difficultness. When he was asked to choose a problem, if he chose the easiest one, Mr. G would complain that “Hey, what a trivial problem!”; if he chose the hardest one, Mr. M would grumble that it took too much time to finish it. To address this dilemma, Mr. B decided to take the one with the medium difficulty. Therefore, he needed a way to know the median number in the given interval of the sequence.

Input

For each test case, the first line contains a single integer n (1 <= n <= 100,000) indicating the total number of problems. The second line contains n integers xi (0 <= xi <= 1,000,000,000), separated by single space, denoting the difficultness of each problem, already sorted by publish time. The next line contains a single integer m (1 <= m <= 100,000), specifying number of queries. Then m lines follow, each line contains a pair of integers, A and B (1 <= A <= B <= n), denoting that Mr. B needed to choose a problem between positions A and B (inclusively, positions are counted from 1). It is guaranteed that the number of items between A and B is odd.

Output

For each query, output a single line containing an integer that denotes the difficultness of the problem that Mr. B should choose.

Sample Input

5
5 3 2 4 1
3
1 3
2 4
3 5
5
10 6 4 8 2
3
1 3
2 4
3 5

Sample Output

Case 1:
3
3
2
Case 2:
6
6
4

题意：

题目给你一个序列n，以及m个查询，问你查询的区间中的中位数。

思路：

可以用主席树和划分树去做，这里用的划分树，就是划分树的模板，但是需要注意的是，在查询区间中位数的时候中位数就是区间的差（r- l）减去头（l）+ 1，然后再套用模板就行了。

#include 
#include 
#include 
#include 
using namespace std;
const int maxn = 1e+5 + 10;
int tree[20][maxn], sorted[maxn], toleft[20][maxn];
void Build(int l, int r, int dep) {
    if (l == r) return ;
    int mid = (l + r) >> 1;
    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) >> 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() {
    int n, m, Case = 1;
    while (scanf("%d", &n) != EOF) {
        memset(tree, 0, sizeof(tree));
        for (int i = 1; i <= n; i++) {
            scanf("%d", &sorted[i]);
            tree[0][i] = sorted[i];
        }
        sort(sorted + 1, sorted + n + 1);
        Build(1, n, 0);
        printf("Case %d:\n", Case++);
        scanf("%d", &m);
        while (m--) {
            int l, r, k;
            scanf("%d %d", &l, &r);
            k = ((r - l) >> 1) + 1;
            printf("%d\n", Query(1, n, l, r, 0, k));
        }
    }
    return 0;
}