poj2104 K-th Number

最新推荐文章于 2020-04-17 14:40:12 发布

原创最新推荐文章于 2020-04-17 14:40:12 发布 · 227 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#acm #poj #数据 #结构

ACM 专栏收录该内容

20 篇文章

订阅专栏

本文针对K-thNumber问题，详细介绍了如何使用数据结构快速找出数组某段区间内第k小的数。面对宏硬公司的数据结构部门挑战，文章提供了一种暴力解法，并暗示了更高效的归并树解法。

K-th Number

Time Limit: 20000MS		Memory Limit: 65536K
Total Submissions: 55738		Accepted: 19177
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

看了一些博客，说的这题要用归并树解，可是我不会啊，写的暴力用了7000MS过了，看了一下这题时间限制是20000MS！！

要是时间限制严格一点暴力肯定过不了。还是得把归并树搞懂啊。

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;

struct node
{
    int v;
    int add;
};
bool cmp(node a,node b)
{
    return a.v<b.v;
}
node a[100005];
int main()
{
    int m,n;
    scanf("%d%d",&m,&n);
    for(int i=1;i<=m;i++)
    {
        scanf("%d",&a[i].v);
        a[i].add=i;
    }
    int p,q,k;
    sort(a+1,a+m+1,cmp);
    while(n--)
    {
        scanf("%d%d%d",&p,&q,&k);
        for(int i=1;i<=m;i++)
        {
            if(a[i].add>=p&&a[i].add<=q)
            {
                k--;
            }
            if(k==0)
            {
                printf("%d\n",a[i].v);
                break;
            }
        }
    }
}