Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8707    Accepted Submission(s): 2928

 

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

题意：就是给你一个串，要求求出当前串最大值减最小值大于等于m小于等于k的最长长度

第一次看这个不是很懂，当初刚学单调队列，不是很懂，就放一放，这次看还行，懂了。我们要求在符合条件下的最长长度，单调队列就是可以求得一个序列最小，最大值，那我们每次对比最大最小值是否符合条件，大于k时要求出来靠前一位的是最大还是最小值，当最大最小差值大于k那就不符合条件了，这时那个靠前的最值之前的都不符合，只能在这个最值的下一位以后的序列里找符合题意的长度。所以now就是记录那个最值的后一位。

#include<stdio.h>
#include<queue>
#include<string.h>
#include<algorithm>
using namespace std;
const int maxn=100005;
int a[maxn],q1[maxn],q2[maxn];
int main()
{
    int n,m,k;
    while(~scanf("%d%d%d",&n,&m,&k))
    {
        for(int i=0; i<n; i++)
            scanf("%d",&a[i]);
        int head1=0,head2=0,tail1=0,tail2=0,now=0,ans=0;
        for(int i=0; i<n; i++)
        {
            while(head1<tail1&&a[q1[tail1-1]]<a[i])
                tail1--;
            while(head2<tail2&&a[q2[tail2-1]]>a[i])
                tail2--;
            q1[tail1++]=q2[tail2++]=i;
            while(head1<tail1&&head2<tail2&&a[q1[head1]]-a[q2[head2]]>k)
            {
                if(q1[head1]<q2[head2])
                    now=q1[head1++]+1;
                else
                    now=q2[head2++]+1;
            }
            if(head1<tail1&&head2<tail2&&a[q1[head1]]-a[q2[head2]]>=m)
                ans=max(ans,i-now+1);
        }
        printf("%d\n",ans);
    }
}