Codeforces Round #305 (Div. 1) B. Mike and Feet

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 strengthof
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 test(s)

input

10
1 2 3 4 5 4 3 2 1 6

output

6 4 4 3 3 2 2 1 1 1

这题可以用单调栈做，维护一个栈，记录minmum(该区间的最小值）和count（区间的总长度）。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define maxn 200060
int ans[maxn];
struct node{
	int count,minmum;
}stack[maxn];
int main()
{
	int n,m,i,j,top,count,b;
	while(scanf("%d",&n)!=EOF)
	{
		memset(ans,0,sizeof(ans));
		top=0;
		for(i=1;i<=n;i++){
			scanf("%d",&b);
			count=0;
			while(top>0 && stack[top].minmum>=b){
				stack[top].count+=count;
				count=stack[top].count;
				if(ans[count]<stack[top].minmum){
					ans[count]=stack[top].minmum;
				}
				top--;
			}
			top++;
			stack[top].minmum=b;
			stack[top].count=count+1;
		}
		count=0;
		while(top>0){
				stack[top].count+=count;
				count=stack[top].count;
				if(ans[count]<stack[top].minmum){
					ans[count]=stack[top].minmum;
				}
				top--;
		}
		for(i=n;i>=2;i--){
			if(ans[i]>ans[i-1]){/*这里算出来的ans[i]是连续长度为i的区间的最小值，但这个最小值是所有连续长度为i的区间长度的最大值，下面如果ans[i+1]比ans[i]大，那么ans[i]可以更新为ans[i+1],因为如果i+1个连续数区间的最小值的最大值是b,那么去掉一个数，一定可以做到长度为i的连续数区间的最大值是b。*/
				ans[i-1]=ans[i];
			}
		}
		for(i=1;i<=n-1;i++){
			printf("%d ",ans[i]);
		}
		printf("%d\n",ans[i]);
	}
	return 0;
}