POJ2104 K-th Number 【归并树】

K-th Number

Time Limit: 20000MS Memory Limit: 65536KTotal Submissions: 38379 Accepted: 12480Case 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.

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 31 5 2 6 3 7 42 5 34 4 11 7 3

Sample Output

563

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

题意：给定一个包含n个不同数的数列a1, a2, ..., an 和m个三元组表示的查询。对于每个查询(i, j, k), 输出ai, ai+1, ... ,aj的升序排列中第k个数 。

题解：用线段树，每个节点维护一个区间并且保证内部升序，对于每次查询x，返回该区间小于x的数的个数。就这样不断二分，直到找到x为止。

#include <stdio.h>#include <string.h>#include <algorithm>#include <vector>#define maxn 100005#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1using namespace std;vector<int> T[maxn << 2];int N, Q;void build(int l, int r, int rt) {    if(l == r) {        int val;        scanf("%d", &val);        T[rt].push_back(val);        return;    }    int mid = (l + r) >> 1;    build(lson);    build(rson);    T[rt].resize(r - l + 1); // Attention    merge(T[rt<<1].begin(), T[rt<<1].end(), T[rt<<1|1].begin(), T[rt<<1|1].end(), T[rt].begin());}int query(int L, int R, int val, int l, int r, int rt) {    if(L == l && R == r) {        return upper_bound(T[rt].begin(), T[rt].end(), val) - T[rt].begin();    }    int mid = (l + r) >> 1;    if(R <= mid) return query(L, R, val, lson);    else if(L > mid) return query(L, R, val, rson);    return query(L, mid, val, lson) + query(mid + 1, R, val, rson);}int main() {    int a, b, c, k, left, right, mid;    scanf("%d%d", &N, &Q);    build(1, N, 1);    while(Q--) {        scanf("%d%d%d", &a, &b, &k);        left = -1; right = N - 1;        while(right - left > 1) { // binary search            mid = (left + right) >> 1;            c = query(a, b, T[1][mid], 1, N, 1);            if(c >= k) right = mid;            else left = mid;        }        printf("%d\n", T[1][right]);    }    return 0;}