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HDU - 3530 Subsequence

2024-07-13 22:44:25   224人阅读

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
 

Source

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

题意：求一个最长的最大和最小的差在[m, k]范围的序列

思路：维护一个最大序列和一个最小序列，然后判断是否符合范围就行了

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <queue>
using namespace std;
const int maxn = 100010;

int num[maxn];
int q1[maxn], q2[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", &num[i]);
		int ans = 0;
		int rear1 = 0, front1 = 0, rear2 = 0, front2 = 0;
		int cnt = 0;
		for (int i = 1; i <= n; i++) {
			while (front1 < rear1 && num[q1[rear1-1]] < num[i])
				rear1--;
			q1[rear1++] = i;
			while (front2 < rear2 && num[q2[rear2-1]] > num[i])
				rear2--;
			q2[rear2++] = i;
			while (front1 < rear1 && front2 < rear2 && num[q1[front1]]-num[q2[front2]] > k) {
				if (q1[front1] < q2[front2]) {
					cnt = q1[front1];
					front1++;
				}else {
					cnt = q2[front2];
					front2++;
				}
			}
			if (front1 < rear1 && front2 < rear2 && num[q1[front1]]-num[q2[front2]] >= m)
				ans = max(ans, i-cnt);
		}
		printf("%d\n", ans);
	}
	return 0;
}


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