文章标题 HDU 3530 : Subsequence（单调队列）

Subsequence

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 1,1000001,100000. m and k are in the range 0,10000000,1000000. The second line has n integers, which are all in the range 0,10000000,1000000.
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：有n个数的序列，然后给出数m,k,然后要我们求出这个n个序列中的最长子序列，使得这个序列满足最大值与最小值的查值大于m且小于k。
分析：用单调队列，用两个队列，一个单调上升，一个单调下降。来维护当前的序列的最大值与最小值，然后当出现不满足的情况的时候应该从那个位置值小的来维护，然后更新答案即可。
代码:

#include<iostream>#include<string>#include<cstdio>#include<cstring>#include<vector>#include<math.h>#include<map>#include<queue> #include<algorithm>using namespace std;const int inf = 0x3f3f3f3f;typedef pair<int,int> pii;int n,m,k; int a[100005];int q1[100005],q2[100005];//q1单调上升队列，q2单调下降队列，保存的是位置 int rear1,rear2;int head1,head2;int main (){    while (scanf ("%d%d%d",&n,&m,&k)!=EOF){        rear1=rear2=0;        head1=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--;//降序             while (head2<rear2&&a[q2[rear2-1]]>a[i])rear2--;//升序            q1[rear1++]=i;            q2[rear2++]=i;            while (head1<rear1&&head2<rear2&&a[q1[head1]]-a[q2[head2]]>k){                //当最值之间的差值超过了k，应该在那个位置小的先移动                 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){                ans=max(ans,i-now+1);//更新答案             }        }        printf ("%d\n",ans);    }    return 0;}