hdu 3530 Subsequence(DP+单调队列优化)

Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2518    Accepted Submission(s): 826

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

   
   
   
   

    
    
    
    5
4
   
   
   
   

 

Source

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

 

Recommend

zhengfeng

 

题目：http://acm.hdu.edu.cn/showproblem.php?pid=3530

题意：给你一个长度为n的数列，要求一个子区间，使得区间的最大值与最小值的差s满足,m<=s<=k，求满足条件的最长子区间

分析：做了前面几题后，这题容易想到用两个单调队列维护当前最值，作为判断条件，如果差值大于k了，就去掉较前面的那个队列元素，并把区间头更新为它的标号+1，这里注意差值小于m并不需要去掉元素，还有更新答案时要先判断是否满足条件才能更新，因为这两条我又wa了n次啊= =

代码：

#include<cstdio>
#include<iostream>
using namespace std;
const int mm=111111;
int a[mm],qa[mm],qb[mm];
int main()
{
    int i,n,m,k,ans,l,la,ra,lb,rb;
    while(~scanf("%d%d%d",&n,&m,&k))
    {
        ans=l=la=lb=0,ra=rb=-1;
        for(i=1;i<=n;++i)
        {
            scanf("%d",&a[i]);
            while(la<=ra&&a[qa[ra]]>a[i])--ra;
            while(lb<=rb&&a[qb[rb]]<a[i])--rb;
            qa[++ra]=i;
            qb[++rb]=i;
            while(a[qb[lb]]-a[qa[la]]>k)
                l=(qa[la]<qb[lb])?qa[la++]:qb[lb++];
            if(a[qb[lb]]-a[qa[la]]>=m&&a[qb[lb]]-a[qa[la]]<=k)ans=max(ans,i-l);
        }
        printf("%d\n",ans);
    }
    return 0;
}