Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2641    Accepted Submission(s): 869

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       单调队列。 用两个单调队列维护最大值和最小值。    

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

const int MAXN=100010;
int q1[MAXN],q2[MAXN];
int rear1,head1;
int rear2,head2;
int a[MAXN];
int main()
{
  //  freopen("in.txt","r",stdin);
  //  freopen("out.txt","w",stdout);
    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]);
            while(head1<rear1&&a[q1[rear1-1]]<a[i])rear1--;//这里的等号取和不取都可以的
            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)
            {
                //int t=min(q1[head1],q2[head2]);
                if(ans<i-now+1)ans=i-now+1;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}