Subsequence

最新推荐文章于 2019-08-20 22:34:09 发布

转载最新推荐文章于 2019-08-20 22:34:09 发布 · 98 阅读

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原文链接：http://www.cnblogs.com/ERKE/p/3574213.html

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#java

本文介绍了一种使用单调队列解决特定子序列问题的方法。该问题要求找出整数序列中最长的子序列，使得子序列中最大元素与最小元素之差不小于m且不大于k。通过双队列维护候选元素，实现高效求解。

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Subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3892 Accepted Submission(s): 1271

Problem 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

单调队列真的有很多奥妙！

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 using namespace std;
 5 int z[110000],a[110000],b[110000];
 6 int main()
 7 {
 8     int n,m,k,i;
 9     while(~scanf("%d%d%d",&n,&m,&k))
10     {
11         for(i=0; i<n; i++)
12             scanf("%d",&z[i]);
13         int top1=0,top2=0,tail1=0,tail2=0,last1=-1,last2=-1,an=0;
14        for(i=0;i<n;i++)
15        {
16            while(tail1>top1&&z[a[tail1-1]]<=z[i])tail1--;
17            a[tail1++]=i;
18            while(tail2>top2&&z[b[tail2-1]]>=z[i])tail2--;
19            b[tail2++]=i;
20            while(z[a[top1]]-z[b[top2]]>k)
21            {
22                if(a[top1]<b[top2])
23                last1=a[top1++];
24                else last2=b[top2++];
25            }
26            if(z[a[top1]]-z[b[top2]]>=m)
27            {
28                an=max(an,i-max(last1,last2));
29            }
30        }
31        cout<<an<<endl;
32     }
33 }