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

B. Mike and Feet

Time Limit: 20 Sec  Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/547/problem/B

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 a_i 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, ..., a_n (1 ≤ a_i ≤ 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

10
1 2 3 4 5 4 3 2 1 6

Sample Output

6 4 4 3 3 2 2 1 1 1 

HINT

题意

给你一个堆数，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：

用一个类似单调栈的思想，处理以这个点为最小值可以往左右延伸多少，然后乱搞一下就好了

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 200001
#define mod 1000000007
#define eps 1e-9
int Num;
char CH[];
//const int inf=0x7fffffff;   //нчоч╢С
const int inf=0x3f3f3f3f;
inline ll read()
{
    ll x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
inline void P(int x)
{
    Num=;if(!x){putchar('');puts("");return;}
    while(x>)CH[++Num]=x%,x/=;
    while(Num)putchar(CH[Num--]+);
    puts("");
}
//**************************************************************************************

int a[maxn];
int dp[maxn];
int l[maxn];
int r[maxn];
int main()
{
    //test;
    int n=read();
    for(int i=;i<=n;i++)
        a[i]=read();
    a[]=-,a[n+]=-;

    for(int i=;i<=n;i++)
    {
        int j=i-;
        while(a[j]>=a[i])j=l[j];
        l[i]=j;
    }
    for(int i=n;i>=;i--)
    {
        int j=i+;
        while(a[j]>=a[i])j=r[j];
        r[i]=j;
    }
    for(int i=;i<=n;i++)
    {
        int len=r[i]-l[i]-;
        dp[len]=max(dp[len],a[i]);
    }
    for(int i=n-;i>=;i--)
        dp[i]=max(dp[i+],dp[i]);
    for(int i=;i<=n;i++)
        cout<<dp[i]<<" ";
}

10
1 2 3 4 5 4 3 2 1 6