hdu 3530 （单调队列）

Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4441    Accepted Submission(s): 1457

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

用两个单调队列维护最大值和最小值

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<cmath>
#include<map>
#include<algorithm>
using namespace std;
int p[101010];
int mmax[101010],mmin[101010];
int main(){
    int n,m,k;
    while(cin>>n>>m>>k){
        for(int i=1;i<=n;i++) scanf("%d",p+i);
        int head1,tail1,head2,tail2,be=1,mm=0;
        head1=tail1=0;
        head2=tail2=0;
        for(int i=1;i<=n;i++){
            while(head1<tail1&&p[mmax[tail1-1]]<p[i]) tail1--;//维护单调递增
            while(head2<tail2&&p[mmin[tail2-1]]>p[i]) tail2--;//维护单调递减
            mmax[tail1++]=i;
            mmin[tail2++]=i;
            while(head1<tail1&&head2<tail2&&p[mmax[head1]]-p[mmin[head2]]>k){
                if(mmax[head1]<mmin[head2]) be=mmax[head1++]+1;
                else be=mmin[head2++]+1;
            }
            if(head1<tail1&&head2<tail2&&p[mmax[head1]]-p[mmin[head2]]>=m){
                mm=max(mm,i-be+1);
            }
        }
        cout<<mm<<endl;
    }
}