HDU3530 单调队列的应用

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

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

/**
HDU 3530  单调队列的应用
题意：
     给定一段序列，求出最长的一段子序列使得该子序列中最大最小只差x满足m<=x<=k。
解题思路：
     建立两个单调队列分别递增和递减维护（头尾删除，只有尾可插入）
     Max - Min 为两个队列的队首之差while(Max-Min>K) 看哪个的队首元素比较前就移动谁的
     最后求长度时，需要先记录上一次的被淘汰的最值位置last ，这样[last+1,i]即为满足条件的连续子序列了
i - last
*/
#include 
#include 
#include 
#include 
using namespace std;
const int N=100005;
int q_max[N],q_min[N];//递增，递减
int a[N],n,m,k;
int main()
{
    while(~scanf("%d%d%d",&n,&m,&k))
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        int head_min=0,head_max=0,tail_min=0,tail_max=0;
        int left1=0,left2=0;
        int maxx=0;
        for(int i=1;i<=n;i++)
        {
            while(head_min=a[i])
                tail_max--;
            q_max[tail_max++]=q_min[tail_min++]=i;
           /* printf("***%d 递减、递增***\n",i);
            for(int j=head_min;jk)
            {
                if(q_min[head_min]=m)
                  maxx=max(maxx,i-max(left1,left2));
        }
        printf("%d\n",maxx);
    }
    return 0;
}
/*
5 2 3
1 -1 2 -6 5
5 1 3
1 2 3 4 5
6 0 0
-1 0 2 1 125 -5
*/