Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

单调队列

链接：HDU 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

题意：

第一行输入三个数：n，a，b；
接下来输入n个数
题目意思是让你在序列a[n]中找出一个最长序列，使其最大值与最小值之差在[a,b]之间，问这个序列最长是多少；

解题思路：

利用两个单调队列，一个求最大值，一个求最小值，最大值-最小值要在[a,b]区间内；

完整代码：

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;
///队列单减：从头到尾依次递减，头大尾小;
///队列单增：从头到尾依次递增，头小尾大;
int f[100010];
int qmax[100010],head1,tail1;///单减队列，维护最大值
int qmin[100010],head2,tail2;///单增队列，维护最小值
int main()
{
    int n,a,b,ans,head;
    while(scanf("%d%d%d",&n,&a,&b)!=EOF)
    {
        memset(f,0,sizeof(f));
        memset(qmax,0,sizeof(qmax));
        memset(qmin,0,sizeof(qmin));
        head1=head2=0;
        tail1=tail2=0;
        ans=0,head=0;
        for(int i=0;i<n;i++)
        {
             scanf("%d",&f[i]);
             while(head1<tail1&&f[i]-f[qmax[tail1-1]]>=0)///利用单减队列维护最大值
                tail1--;
             while(head2<tail2&&f[i]-f[qmin[tail2-1]]<=0)///利用单增队列维护最小值
                tail2--;
             qmax[tail1]=i; qmin[tail2]=i;
             tail1++; tail2++;
             while(head1<tail1&&head2<tail2&&f[qmax[head1]]-f[qmin[head2]]>b)///他们的差大于所给的值后
             {
                 ///用一个head来记录目前满足该条件的头的值
                if(qmax[head1]<qmin[head2])
                  {
                      head=qmax[head1];
                      head1++;
                  }
                else
                  {
                      head=qmin[head2];
                      head2++;
                  }
                  head++;
             }
             if(head1<tail1&&head2<tail2&&f[qmax[head1]]-f[qmin[head2]]>=a)///他们的差>=所给的值时，求ans
             {
                ans=max(ans,i-head+1);///要求最长的序列
             }
        }
        printf("%d\n",ans);
    }
    return 0;
}