HDU 3530 Subsequences(单调队列)

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解题思路：

开两个单调队列即可。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#define LL long long 
using namespace std;
const int maxn = 100000 + 10;
int A[maxn];
int Q1[maxn];
int Q2[maxn];
int P1[maxn];
int P2[maxn];
int Min[maxn];
int Max[maxn];
int N, M, K;
int main()
{
	while(scanf("%d%d%d", &N, &M, &K)!=EOF)
	{
		for(int i=1;i<=N;i++)
			scanf("%d", &A[i]);
		int head1 = 1, tail1 = 0;
		int head2 = 1, tail2 = 0;
		int now = 0, ans = 0;
		for(int i=1;i<=N;i++)
		{
			while(head1 <= tail1 && A[P1[tail1]] <= A[i])
				tail1--;
			P1[++tail1] = i;
			while(head2 <= tail2 && A[P2[tail2]] >= A[i])
				tail2--;
			P2[++tail2] = i;
			while(A[P1[head1]] - A[P2[head2]] > K)
			{
				if(P1[head1] < P2[head2])
					now = P1[head1++];
				else
					now = P2[head2++];
			}
			if(A[P1[head1]] - A[P2[head2]] >= M)
				ans = max(ans, i - now);

		}
		printf("%d\n", ans);
	}
	return 0;
}

HDU 3530 Subsequences(单调队列)

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原文地址：http://blog.csdn.net/moguxiaozhe/article/details/43567299