K-th Number(第k大数)

最新推荐文章于 2020-09-12 12:23:19 发布

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最新推荐文章于 2020-09-12 12:23:19 发布

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文章标签： java 数据结构与算法

本文深入探讨了如何使用排序和位置记录来高效解决K-thNumber问题，即快速找到数组区间内的第k小元素。通过实例分析和代码实现，详细解释了解决方案的原理与应用。

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K-th Number

Time Limit: 20000MS		Memory Limit: 65536K
Total Submissions: 45710		Accepted: 15199
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

Hint

This problem has huge input,so please use c-style input(scanf,printf),or you may got time limit exceed.

题解：初看这个题，完全就感觉是神题，想不出思路，不敢暴力，其实哪有那么麻烦，可以先对序列排序，记录下其位置就好；然后对于每个询问，从1到n遍历一遍；看id在不在l,r之间；如果在就t--;当t等于0的时候，相当于找到了这个第k大的数，这就是答案；还有的大神用主席树，函数线段树，划分树什么的写的。。。。。我能说什么吗

代码：

extern "C++"{
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
using namespace std;
typedef long long LL;
void SI(int &x){scanf("%d",&x);}
void SI(double &x){scanf("%lf",&x);}
void SI(char *x){scanf("%s",x);}
//void SI(LL &x){scanf("%lld",&x);}

void PI(int &x){printf("%d",x);}
void PI(double &x){printf("%lf",x);}
void PI(char *x){printf("%s",x);}
//void PI(LL &x){printf("%lld",x);}

}
const int MAXN = 100010;
struct Node{
    int id,v;
    bool operator < (const Node &b)const{
        return v < b.v;
    }
};
Node dt[MAXN];
int main(){
    int n,m;
    while(~scanf("%d%d",&n,&m)){
        for(int i = 1;i <= n;i++){
            SI(dt[i].v);
            dt[i].id = i;
        }
        sort(dt + 1,dt + n + 1);
        //for(int i = 1;i <= n;i++)printf("%d ",dt[i].v);puts("");
        int l,r,t;
        while(m--){
            scanf("%d%d%d",&l,&r,&t);
            for(int i = 1;i <= n;i++){
                if(dt[i].id >= l && dt[i].id <= r){
                    t--;
                    if(t == 0){
                        printf("%d\n",dt[i].v);break;
                    }
                }
            }
        }
    }
    return 0;
}

java:

package com.lanqiao.week1;

import java.lang.reflect.Array;
import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;

public class poj2104 {
    private static Scanner cin;
    private static int MOD = 1000000007;
    static{
        cin = new Scanner(System.in);
    }
    private static class Node{
        int i, v;
    }
    public static void main(String[] args) {
        int n, m;
        n = cin.nextInt();
        m = cin.nextInt();
        Node[] arr = new Node[n];
        for(int i = 0; i < n; i++){
            arr[i] = new Node();
            arr[i].i = i + 1;
            arr[i].v = cin.nextInt();
        }
        Arrays.sort(arr, new Comparator<Node>() {

            @Override
            public int compare(Node a, Node b) {
                if(a.v != b.v){
                    return a.v - b.v;
                }else{
                    return a.i - b.i;
                }
            }
            
        });
        while(m-- > 0){
            int s, e, k;
            s = cin.nextInt();
            e = cin.nextInt();
            k = cin.nextInt();
            int cnt = 0;
            for(int i = 0; i < n ; i++){
                if(arr[i].i >= s && arr[i].i <= e){
                    cnt ++;
                    if(cnt == k){
                        System.out.println(arr[i].v);
                        break;
                    }
                }
            }
        }
    }
    
}