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HDU 3530 Subsequence

[编程题]HDU 3530 Subsequence

热度指数：88 时间限制：C/C++ 1秒，其他语言2秒空间限制：C/C++ 32M，其他语言64M
算法知识视频讲解

There is a sequence of integers. Your task is to find the longest subsequence that satisfies the following condition: the difference between the maximum element and the minimum element of the subsequence is no smaller than m and no larger than k.

输入描述:

There are multiple test cases.
       
For each test case, the first line has three integers, n, m and k. n is the length of the sequence and is in the range [1, 100000]. m and k are in the range [0, 1000000]. The second line has n integers, which are all in the range [0, 1000000].
       
Proceed to the end of file.

输出描述:

For each test case, print the length of the subsequence on a single line.

示例1

输入

输出

5
4

算法知识视频讲解

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Prelacyso

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;

const int MAXN=100010;
int q1[MAXN],q2[MAXN];
int rear1,head1;
int rear2,head2;
int a[MAXN];
int main()
{
    int n,m,k;

    while(scanf("%d%d%d",&n,&m,&k)!=EOF)
    {
        rear1=head1=0;
        rear2=head2=0;
        int ans=0;
        int now=1;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            while(head1<rear1&&a[q1[rear1-1]]<a[i])rear1--;
            while(head2<rear2&&a[q2[rear2-1]]>a[i])rear2--;
            q1[rear1++]=i;
            q2[rear2++]=i;
            while(head1<rear1&&head2<rear2&&a[q1[head1]]-a[q2[head2]]>k)
            {
                if(q1[head1]<q2[head2])now=q1[head1++]+1;
                else now=q2[head2++]+1;
            }
            if(head1<rear1&&head2<rear2&&a[q1[head1]]-a[q2[head2]]>=m)
            {
                if(ans<i-now+1)ans=i-now+1;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}

发表于 2015-10-28 15:17:57 回复(0)

fengzhk

import java.util.LinkedList;
import java.util.Scanner;
public class Main {
	public static int LongestDiffLength(int[] arr,int m,int k) {
		if(arr == null || arr.length == 0)
			return 0;
		
		LinkedList<Integer> qmin = new LinkedList<Integer>();
		LinkedList<Integer> qmax = new LinkedList<Integer>();
		int i = 0;
		int start = 0;
		int max = 0;
		while(i < arr.length) {
			
			while(!qmin.isEmpty() && arr[qmin.peekLast()] > arr[i]) {
				qmin.pollLast();
			}
			qmin.addLast(i);
			while(!qmax.isEmpty() && arr[qmax.peekLast()] < arr[i]) {
				qmax.pollLast();
			}
			qmax.addLast(i);
			
			while(arr[qmax.getFirst()] - arr[qmin.getFirst()] > k) {
				if (qmax.getFirst() < qmin.getFirst()) {
					start = qmax.peekFirst() + 1;
					qmax.pollFirst();
				}else {
					start = qmin.peekFirst() + 1;
					qmin.pollFirst();
				}
			}
			if(arr[qmax.getFirst()] - arr[qmin.getFirst()] >= m) {
				max = Math.max(max, i - start + 1);
			}
			i++;
				
		}
		
		return max;
		
	}
	
	
	public static void main(String[] args) {
		int n;
		int m;
		int k;
		Scanner sc = new Scanner(System.in);
		while (sc.hasNext()) {
			n = sc.nextInt();
			int[] Arr = new int[n];
			m = sc.nextInt();
			k = sc.nextInt();
			for(int i = 0; i < n;i++) {
				Arr[i] = sc.nextInt();
			}
		System.out.println(LongestDiffLength(Arr,m,k));
		}
	}


}

编辑于 2018-02-26 23:31:37 回复(1)