HDU 3530 Subsequence(单调队列)
Subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2641 Accepted Submission(s): 869
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.
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
Recommend
zhengfeng
单调队列。
用两个单调队列维护最大值和最小值。
#include<stdio.h> #include<iostream> #include<algorithm> #include<string.h> using namespace std; const int MAXN=100010; int q1[MAXN],q2[MAXN]; int rear1,head1; int rear2,head2; int a[MAXN]; int main() { // freopen("in.txt","r",stdin); // freopen("out.txt","w",stdout); int n,m,k; while(scanf("%d%d%d",&n,&m,&k)!=EOF) { rear1=head1=0; rear2=head2=0; int ans=0; int now=1; for(int i=1;i<=n;i++) { scanf("%d",&a[i]); while(head1<rear1&&a[q1[rear1-1]]<a[i])rear1--;//这里的等号取和不取都可以的 while(head2<rear2&&a[q2[rear2-1]]>a[i])rear2--; q1[rear1++]=i; q2[rear2++]=i; while(head1<rear1&&head2<rear2&&a[q1[head1]]-a[q2[head2]]>k) { if(q1[head1]<q2[head2])now=q1[head1++]+1; else now=q2[head2++]+1; } if(head1<rear1&&head2<rear2&&a[q1[head1]]-a[q2[head2]]>=m) { //int t=min(q1[head1],q2[head2]); if(ans<i-now+1)ans=i-now+1; } } printf("%d\n",ans); } return 0; }
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