hdu3530 Subsequence

最新推荐文章于 2019-11-19 21:01:08 发布

原创最新推荐文章于 2019-11-19 21:01:08 发布 · 2k 阅读

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文章标签：

#training #each #output #ini #input #2010

此博客讨论了如何在给定序列中找到一个子序列，使得该子序列中的最大值与最小值之差不小于m且不大于k。通过使用单调队列，有效地解决了这一问题，并提供了实例输入和输出来说明解决方案。

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Subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2136 Accepted Submission(s): 708

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

Sample Output

5
4

Source

2010 ACM-ICPC Multi-University Training Contest（10）——Host by HEU

Recommend

zhengfeng

单调队列。

维护最大值和最小值，如果发现最大值和最小值的差大于k，那么就移动下标最靠前的队列。

注意如下数据：

5 2 4

2 1 5 2 2

应该用一个last标记上一个移动的位置，然后答案就是max{i-last}，之前没有这个标记wa了一次。

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

int a[100005];
deque <int> q1;
deque <int> q2;

int main()
{
    int i,j,n,x,y,ans,last;
    while(scanf("%d%d%d",&n,&x,&y)!=EOF)
    {
        for (i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        ans=0;
        last=-1;
        while(!q1.empty()) q1.pop_front();
        while(!q2.empty()) q2.pop_front();
        for (i=0;i<n;i++)
        {
            while(!q1.empty() && a[i]>a[q1.back()]) q1.pop_back();
            while(!q2.empty() && a[i]<a[q2.back()]) q2.pop_back();
            q1.push_back(i);
            q2.push_back(i);
            while(!q1.empty() && !q2.empty() && a[q1.front()]-a[q2.front()]>y)
            {
                if (q1.front()>q2.front())
                {
                    last=q2.front();
                    q2.pop_front();
                }
                else if (q1.front()<q2.front())
                {
                    last=q1.front();
                    q1.pop_front();
                }
                else
                {
                    last=q1.front();
                    q1.pop_front();
                    q2.pop_front();
                }
            }
            if (!q1.empty() && !q2.empty() && a[q1.front()]-a[q2.front()]>=x)
            {
                ans=max(ans,i-last);
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}