HDU 3530 --- Subsequence 单调队列

HDU 3530  --- Subsequence 单调队列

Description

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.

 

Input

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.

 

Output

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

 

Sample Input

5 0 0

1 1 1 1 1

5 0 3

1 2 3 4 5

 

Sample Output

5 4

 

 

释意：

　　第一行给出三个数n、m、k，第二行给出一个n个数序列，要求找出一个最长的连续子序列，要求子序列的最大值与最小值的差值要小于等于k大于等于m。

　　输出满足该条件的最长子序列的长度。

题解：

　　由于要求最大值与最小值，所以建立两个单调队列，一个单调递增队列，一个单调递减队列，从而通过队头确定当前子序列的最大值与最小值，我们可以仅以

最大值-最小值 是否大于k来确定满足的当前序列的子序列，至于是否满足大于等于买这个条件，可以在比较时来确定是否加入比较的行列。

　　对于如何确定当前序列的满足条件的最长子序列问题：

1.求两队对头的差值，若小于等于k则满足。如不满足则将两队中队头元素下标较小的一个出队，用last1，last2分别几下当前队头元素的前一个元素的坐标，继续进行队头元素的比较。最终用 当前元素下标 i-max（last1，last2）便是满足条件的最长子序列的长度。

 

代码实现：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
struct Node
{
    int value;
    int position;
}q1[100050],q2[100050]; // q1模拟单调递减队列，q2单调递增队列
int a[100050];
int main()
{
    int n,m,k;
    while(~scanf("%d %d %d",&n,&m,&k))
    {
        int head1 = 1, head2 = 1;
        int tail1 = 0, tail2 = 0;
        int last1 = 0, last2 = 0;
        int ans = 0;  // 一定要初始化为0！！！！！！！
        for(int i = 1;i<=n;i++) scanf("%d",a+i);
        for(int i = 1;i<=n;i++)
        {
            while(head1<=tail1&&q1[tail1].value<=a[i]) tail1--;  //当前元素进队
            q1[++tail1].position = i;
            q1[tail1].value = a[i];

            while(head2<=tail2&&q2[tail2].value>=a[i]) tail2--; //当前元素进队
            q2[++tail2].position = i;
            q2[tail2].value = a[i];
            
            //确定当前序列的最长满足条件的子序列
            
            while(head1<=tail1&&head2<=tail2&&q1[head1].value-q2[head2].value>k)
            {
                if(q1[head1].position<q2[head2].position)
                {
                    last1 = q1[head1++].position;
                }
                else last2 = q2[head2++].position;
            }

            if(q1[head1].value-q2[head2].value>=m)
            ans = max(ans , i-max(last1,last2));  //  细细理解
        }
        cout<<ans<<endl;
    }
    return 0;
}

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