POJ-----2104---K-th Number---暴力

最新推荐文章于 2021-01-11 23:10:18 发布

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本文介绍了一个数据结构问题：快速找出数组某段区间内的第k大数值。通过给出具体实例和样例输入输出，展示了如何使用排序算法直接解决该问题。

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K-th Number

Time Limit: 20000MS		Memory Limit: 65536K
Total Submissions: 53875		Accepted: 18527
Case 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 3
1 5 2 6 3 7 4
2 5 3
4 4 1
1 7 3

Sample Output

5
6
3

给一个数列，问某一段中第k大的值，因为时间比较富裕，直接sort暴力了

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#define LL long long
using namespace std;
const int mod = 1e9+7;
const int maxn = 1e5+10;
struct node{
	int val, id;
}s[maxn];
bool cmp(const node& a, const node& b){
	return a.val < b.val;
}
int main(){
	int m, n, a, b, c;
	cin >> n >> m;
	for(int i = 1; i <= n; i++){
		scanf("%d", &s[i].val);
		s[i].id = i;
	}
	sort(s+1, s+n+1, cmp);
	while(m--){
		scanf("%d%d%d", &a, &b, &c);
		for(int i = 1; i <= n; i++){
			if(s[i].id >= a && s[i].id <= b){
				if(--c == 0){
					printf("%d\n", s[i].val);
					break;
				}
			}
		}
	}
	return 0;
}