Subsequence（HDU-3530）（单调队列）

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：让我们找到一个最长子序列，使得这个子序列中的最大值和最小值相差不超过k不小于m。

思路：通过大佬的思路发现是维护了两个队列，一个是以当前结束所构成的递减序列的位置，另一个是以当前结束构成的递增序列的位置，然后每次的最大值减去最小值，如果大于k，那么就更新两个中的一个，应该更新位置较小的那个，这样才能使得这个区间的长度最大，然后就这么更新就行了 。

AC代码：

#include 
const int maxx=100010;
const int inf=0x3f3f3f3f;
using namespace std;
int a[maxx],b[maxx],c[maxx];
int main()
{
    int n,m,k;
    while(~scanf("%d%d%d",&n,&m,&k))
    {
        for(int i=1; i<=n; i++)
            scanf("%d",&a[i]);
        int liftl=0,liftt=0,rightr=0,rightt=0,ans=0,pos=0;
        for(int i=1; i<=n; i++)
        {
            while(liftla[i])
                rightt--;
            b[rightr++]=i;
            c[rightt++]=i;
            while(liftlk)
            {
                if(b[liftl]=m)
                ans=max(ans,i-pos);
        }
        printf("%d\n",ans);
    }
    return 0;
}