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B. Mike and Feet
阅读量:5377 次
发布时间:2019-06-15

本文共 1975 字,大约阅读时间需要 6 分钟。

 

B. Mike and Feet

time limit per test

1 second

memory limit per test

256 megabytes
 

Mike is the president of country What-The-Fatherland. There are n bears living in this country besides Mike. All of them are standing in a line and they are numbered from 1 to n from left to right. i-th bear is exactly ai feet high.

A group of bears is a non-empty contiguous segment of the line. The size of a group is the number of bears in that group. The strength of a group is the minimum height of the bear in that group.

Mike is a curious to know for each x such that 1 ≤ x ≤ n the maximum strength among all groups of size x.

Input

The first line of input contains integer n (1 ≤ n ≤ 2 × 105), the number of bears.

The second line contains n integers separated by space, a1, a2, ..., an (1 ≤ ai ≤ 109), heights of bears.

Output

Print n integers in one line. For each x from 1 to n, print the maximum strength among all groups of size x.

input

10 1 2 3 4 5 4 3 2 1 6

output

6 4 4 3 3 2 2 1 1 1

 Solution

题意:给你一个长度为N的序列, 对于(1,n)长度,让你找到线段的最小值的最大值是多少

题解:我们处理出L,R两个数组,表示一个数字为线段最小值时的最右边界和最左边界,然后DP一下就好了.

 

1 #include
2 using namespace std; 3 const int maxn = 200000; 4 int n,a[maxn + 99],f[maxn + 99],l[maxn + 99],r[maxn + 99]; 5 int main(){ 6 scanf("%d",&n); 7 for (int i = 1 ; i <= n ; ++i) scanf("%d",&a[i]); 8 a[0] = -1; a[n + 1] = -1; 9 for (int i = 1 ; i <= n ; ++i){10 int j = i - 1;11 while (a[i] <= a[j]) j = l[j];12 l[i] = j;13 }14 for (int i = n ; i >= 1 ; --i){15 int j = i + 1;16 while (a[i] <= a[j]) j = r[j];17 r[i] = j;18 }19 for (int i = 1 ; i <= n ; ++i) f[r[i] - l[i] - 1] = max(f[r[i] - l[i] - 1],a[i]);20 for (int i = n - 1 ; i >= 1 ; --i) f[i] = max(f[i],f[i+1]);21 for (int i = 1 ; i <= n ; ++i) cout<
<<" ";22 return 0;23 }
蒟蒻的代码啦啦啦

 

转载于:https://www.cnblogs.com/juruohx/p/7608487.html

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