BZOJ2096: [Poi2010]Pilots

2096: [Poi2010]Pilots

Time Limit: 30 Sec  Memory Limit: 162 MB
Submit: 914  Solved: 467
[Submit][Status][Discuss]

Description

Tz又耍畸形了!!他要当飞行员,他拿到了一个飞行员测试难度序列,他设定了一个难度差的最大值,在序列中他想找到一个最长的子串,任意两个难度差不会超过他设定的最大值。耍畸形一个人是不行的,于是他找到了你。

Input

输 入:第一行两个有空格隔开的整数k(0<=k<=2000,000,000),n(1<=n<=3000,000),k代表Tz 设定的最大值,n代表难度序列的长度。第二行为n个由空格隔开的整数ai(1<=ai<=2000,000,000),表示难度序列。

Output

输出:最大的字串长度。

Sample Input

3 9
5 1 3 5 8 6 6 9 10

Sample Output

4
(有两个子串的长度为4: 5, 8, 6, 6 和8, 6, 6, 9.最长子串的长度就是4)

HINT

Source

【题解】

单调队列 + 尺取法裸题

 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #define max(a, b) ((a) > (b) ? (a) : (b))
 6 
 7 inline void read(long long &x)
 8 {
 9     char ch = getchar(), c = ch;x = 0;
10     while(ch < '0' || ch > '9')c = ch, ch = getchar();
11     while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar();
12     if(c == '-')x = -x;
13 }
14 
15 const int INF = 0x3f3f3f3f;
16 const int MAXN = 3000000 + 10;
17 
18 long long k,l,n,num,ans,q1[MAXN],q2[MAXN],rank1[MAXN],rank2[MAXN],head1,tail1,head2,tail2;
19 
20 int main()
21 {
22     read(k), read(n);
23     ans = 1;
24     read(num);
25     head1 = tail1 = head2 = tail2 = l = 1, q1[1] = q2[1] = num;
26     rank1[1] = rank2[1] = 1;
27     //q1单调递减
28     //q2单调递增 
29     for(register int i = 2;i <= n;++ i)
30     {
31         read(num);
32         while(num >= q1[tail1] && tail1 >= head1)-- tail1;
33         q1[++tail1] = num;
34         rank1[tail1] = i;
35         while(num <= q2[tail2] && tail2 >= head2)-- tail2;
36         q2[++tail2] = num;
37         rank2[tail2] = i;
38         while(q1[head1] - q2[head2] > k && head1 <= tail1 && head2 <= tail2) 
39         {
40             ++ l;
41             while(rank1[head1] < l && head1 <= tail1)++ head1;
42             while(rank2[head2] < l && head2 <= tail2)++ head2;
43         }
44         ans = max(ans, i - l + 1);
45     }
46     printf("%d", ans);
47     return 0;
48 }
BZOJ2096

 

posted @ 2017-09-03 16:10  嘒彼小星  阅读(142)  评论(0编辑  收藏  举报