HDOJ 3530 Subsequence

Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3501    Accepted Submission(s): 1131

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

 

Source

2010 ACM-ICPC Multi-University Training Contest（10）——Host by HEU

 

Recommend

zhengfeng

 

两个单调队列，如果差值小于m就继续往里面加值，如果差值大于k，那就删掉排在前面的最值。。。。。。。

#include <iostream>

#include <cstdio>

#include <cstring>

using namespace std;

const int maxn=100100;

int dmin[maxn],dmax[maxn];

int f[maxn];

int main()

{

    int n,m,k;

while(scanf("%d%d%d",&n,&m,&k)!=EOF)

{

    int i,ans,minhead,maxhead,mintail,maxtail;

    int st;

    ans=minhead=maxhead=st=0;

    mintail=maxtail=-1;

    for(i=1;i<=n;i++)

    {

        scanf("%d",&f);

        while(maxhead<=maxtail&&f>f[dmax[maxtail]])  maxtail--;

        dmax[++maxtail]=i;

        while(minhead<=mintail&&f<f[dmin[mintail]])  mintail--;

        dmin[++mintail]=i;

        while(f[dmax[maxhead]]-f[dmin[minhead]]>k)

        {

            if(dmax[maxhead]<=dmin[minhead])

            {

                st=dmax[maxhead];

                maxhead++;

            }

            else

            {

                st=dmin[minhead];

                minhead++;

            }

        }

        if(f[dmax[maxhead]]-f[dmin[minhead]]>=m)

        {

            ans=max(ans,i-st);

        }

    }

    printf("%d ",ans);

}

    return 0;

}