HDU3530

1.题目描述：

Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6671    Accepted Submission(s): 2238

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 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

 

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2.题意概述：

给出一个大小为n的数组a[n];

求其中最大值减最小值在[m,k]区间中的字串最长长度。

3.解题思路：

同样可以考虑在从左扫描的同时用单调栈维护当前最值，如果最值超过了就去掉队列前面的元素，至于是去最大值还是最小值应该贪心地去掉下标最小的那个元素。

去除得到合法解，如果区间值符合，就更新一下最值，注意初始化ans应该是0！

4.AC代码：

#include <cstdio>#include <iostream>#include <cstring>#include <string>#include <algorithm>#include <functional>#include <cmath>#include <vector>#include <queue>#include <deque>#include <stack>#include <map>#include <set>#include <ctime>#define INF 0x3f3f3f3f#define maxn 100100#define lson root << 1#define rson root << 1 | 1#define lent (t[root].r - t[root].l + 1)#define lenl (t[lson].r - t[lson].l + 1)#define lenr (t[rson].r - t[rson].l + 1)#define N 1111#define eps 1e-6#define pi acos(-1.0)#define e exp(1.0)using namespace std;const int mod = 1e9 + 7;typedef long long ll;int a[maxn], qmin[maxn], qmax[maxn];int main(){#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);    freopen("out.txt", "w", stdout);    long _begin_time = clock();#endif    int n, m, k;    while (~scanf("%d%d%d", &n, &m, &k))    {        int ans = 0, head1 = 0, head2 = 0, tail1 = 0, tail2 = 0, pre = 0;        a[0] = qmax[0] = qmin[0] = 0;        for (int i = 1; i <= n; i++)        {            scanf("%d", &a[i]);            //维护单调递减            while (head1 <= tail1 && a[qmin[tail1]] > a[i])                tail1--;            //维护单调递增            while (head2 <= tail2 && a[qmax[tail2]] < a[i])                tail2--;            qmin[++tail1] = i;            qmax[++tail2] = i;            while (a[qmax[head2]] - a[qmin[head1]] > k)                pre = qmin[head1] < qmax[head2] ? qmin[head1++] : qmax[head2++];            if (a[qmax[head2]] - a[qmin[head1]] >= m && a[qmax[head2]] - a[qmin[head1]] <= k)                ans = max(ans, i - pre);        }        printf("%d\n", ans);    }#ifndef ONLINE_JUDGE    long _end_time = clock();    printf("time = %ld ms.", _end_time - _begin_time);#endif    return 0;}