CodeForces Round #305 (div1) B. Mike and Feet （单调栈）

题目

本Markdown编辑器使用[StackEdit][6]修改而来，用它写博客，将会带来全新的体验哦：

B - Mike and Feet

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit

Status

Description

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.

Sample Input

Input

10

1 2 3 4 5 4 3 2 1 6

Output

6 4 4 3 3 2 2 1 1 1

看了一个题解才明白：下面链接

这里写链接内容



一开始从头推暴力写法，利用线段书的思想一层层更新取最大值，果断超时了，后来从答案出发想解题

#include<cstdio>
#include<iostream>
#define INT_MAX 2147483647
using namespace std;

int n;
int a[200005];
int l[200005],r[200005];
int ans[200005];
int main(){
cin>>n;
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
a[0]=a[n+1]-INT_MAX;
for(int i=1;i<=n;i++){
int j=i-1;
while(a[j]>=a[i])j=l[j];//跳跃更新//j--;
l[i]=j;
}
for(int i=n;i>0;i--){
int j=i+1;
while(a[j]>=a[i])j=r[j];
r[i]=j;
}
for(int i=1;i<=n;i++){
int d=r[i]-l[i]-1;
if(a[i]>ans[d])ans[d]=a[i];
}
for(int i=n;i>1;i--){
if(ans[i]>ans[i-1])ans[i-1]=ans[i];
}
for(int i=1;i<=n;i++){
printf("%d ",ans[i]);
}
return 0;
}

恩，后续补一波单调栈，希望自己能记得