hdoj-3530-Subsequence

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

构造两个队列来维护当前的最大值和最小值

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;

const int MAXN=100010;
int q1[MAXN],q2[MAXN];
int rear1,head1;
int rear2,head2;
int a[MAXN];
int main()
{
int n,m,k;
while(scanf("%d%d%d",&n,&m,&k)!=EOF)
{
rear1=head1=0;
rear2=head2=0;
int ans=0;
int now=1;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
for(int i=1;i<=n;i++)
{
while(head1<rear1&&a[q1[rear1-1]]<a[i])rear1--;
while(head2<rear2&&a[q2[rear2-1]]>a[i])rear2--;
q1[rear1++]=i;
q2[rear2++]=i;
while(head1<rear1&&head2<rear2&&a[q1[head1]]-a[q2[head2]]>k)
{
if(q1[head1]<q2[head2])now=q1[head1++]+1;
else now=q2[head2++]+1;
}
if(head1<rear1&&head2<rear2&&a[q1[head1]]-a[q2[head2]]>=m)
{
if(ans<i-now+1)ans=i-now+1;
}
}
printf("%d\n",ans);
}
return 0;
}