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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; }}
hdu 3530 （单调队列）
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