Subsequence - HDU 3530 单调队列优化

Subsequence

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

Total Submission(s): 4917 Accepted Submission(s): 1622



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

题意：找最长的连续区间长度，使得最大值与最小值的差不小于m，不大于k。

思路：首先最大值和最小值的差会随着区间的增大而递增的，所以我们一开始让起点为1，找最远的使得最大值与最小值的差<=k的，超过k之后，我们就把起始点往后移动一个，在继续找。这里用到了单调队列来优化一段区间的最大值和最小值。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct node
{
    int num,pos;
}qu1[100010],qu2[100010],temp;
int T,t,n,m,K,num[100010];

int main()
{
    int i,j,k,ans,head1,head2,tail1,tail2,maxn,minn;
    while(~scanf("%d%d%d",&n,&m,&K))
    {
        for(i=1;i<=n;i++)
           scanf("%d",&num[i]);
        head1=head2=1;
        tail1=tail2=0;
        j=1;
        ans=0;
        for(i=1;i<=n;i++)
        {
            temp.num=num[i];
            temp.pos=i;
            while(head1<=tail1 && qu1[tail1].num<=num[i])
              tail1--;
            qu1[++tail1]=temp;
            while(head2<=tail2 && qu2[tail2].num>=num[i])
              tail2--;
            qu2[++tail2]=temp;

            while(true)
            {
                while(head1<tail1 && qu1[head1].pos<j)
                   head1++;
                while(head2<tail2 && qu2[head2].pos<j)
                   head2++;
                maxn=qu1[head1].num;
                minn=qu2[head2].num;
                if(maxn-minn<m)
                  break;
                else if(maxn-minn<=K)
                {
                    ans=max(ans,i-j+1);
                    break;
                }
                else
                  j++;
            }
        }
        printf("%d\n",ans);
    }
}