Subsequence

                                       Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3892    Accepted Submission(s): 1271

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

 单调队列真的有很多奥妙！

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
int z[110000],a[110000],b[110000];
int main()
{
    int n,m,k,i;
    while(~scanf("%d%d%d",&n,&m,&k))
    {
        for(i=0; i<n; i++)
            scanf("%d",&z[i]);
        int top1=0,top2=0,tail1=0,tail2=0,last1=-1,last2=-1,an=0;
       for(i=0;i<n;i++)
       {
           while(tail1>top1&&z[a[tail1-1]]<=z[i])tail1--;
           a[tail1++]=i;
           while(tail2>top2&&z[b[tail2-1]]>=z[i])tail2--;
           b[tail2++]=i;
           while(z[a[top1]]-z[b[top2]]>k)
           {
               if(a[top1]<b[top2])
               last1=a[top1++];
               else last2=b[top2++];
           }
           if(z[a[top1]]-z[b[top2]]>=m)
           {
               an=max(an,i-max(last1,last2));
           }
       }
       cout<<an<<endl;
    }
}

 

 

Source

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