HDU - 3530 Subsequence

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 01 1 1 1 15 0 31 2 3 4 5 

 

Sample Output

 54 

 

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;}