hdu3530 单调队列

最新推荐文章于 2021-01-22 19:59:36 发布

墨碎江南

最新推荐文章于 2021-01-22 19:59:36 发布

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分类专栏：单调队列

本文链接：https://blog.csdn.net/dsaghjkye/article/details/81095304

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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

Sample Output

5
4

题意：给出一个大小为n的数组a[n]; 求其中最大值减最小值在【m,k】中的字串最长长度。

解：开两个队列 s ，一个存下降子序列s1 ，一个存上升子序列s2，那么top代表最值


#include<stdio.h>
#include<string.h>
#include<iostream>
#define maxn 100000+10
using namespace std;
int a[maxn];
int s1[maxn],s2[maxn];
int main()
{
    int n,m,k;
    while(~scanf("%d%d%d",&n,&m,&k))
    {
        int top1,top2,tail1,tail2,last1,last2,ans;
        ans=top1=top2=tail1=tail2=last1=last2=0;
        memset(s1,0,sizeof(s1));
        memset(s2,0,sizeof(s2));
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        for(int i=1;i<=n;i++)
        {
            while(top1<tail1&&a[s1[tail1-1]]<=a[i]) tail1--;//维护下降序列 

            s1[tail1++]=i;

            while(top2<tail2&&a[s2[tail2-1]]>=a[i]) tail2--;//维护上升序列

            s2[tail2++]=i;

            while(a[s1[top1]]-a[s2[top2]]>k)//如果最值之差大于k，那么更新离i最远的位置，让距离尽量最大
            {
               if(s1[top1]<s2[top2])
                    last1=s1[top1++];
               else  last2=s2[top2++];
            }
            if(a[s1[top1]]-a[s2[top2]]>=m)
                ans=max(ans,i-max(last1,last2));//得保证一定满足之差小于等于k

        }
        printf("%d\n",ans);
    }
}