CodeForces 548D 单调栈

Mike and Feet

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

Submit Status Practice CodeForces 548D

Appoint description: 

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

Input

10
1 2 3 4 5 4 3 2 1 6

Output

6 4 4 3 3 2 2 1 1 1 

 

题意:

n个数，现将这n个数中取出连续的k个（k 从1~n）组成一组，组内的值为这一组的最小值，现在要求所有大小为k的组的最大值。

思路:

单调栈。维护一个单调递增栈（从栈底到栈顶单调递增），存每个点贡献的长度和他的值。

 

比如1 2 3 4 5 4 3 2 1 6

 

第一个1，栈为空，那么把它加入栈中，贡献的长度为1，值为1。 

依次直到第6个数 4，由于栈顶元素比4大，那么弹出栈顶元素，并且计算现在的长度（统计现在长度），即弹出的这些

数字能够贡献的长度，知道栈顶元素比当前的值小，然后将当前的4入栈，它能贡献的长度就是统计的长度。

依次类推，要注意最后栈不一定是空的，所以剩下的元素也要处理。

 

/*
 * Author:  sweat123
 * Created Time:  2016/7/12 10:39:09
 * File Name: main.cpp
 */
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<string>
#include<vector>
#include<cstdio>
#include<time.h>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 1<<30
#define MOD 1000000007
#define ll long long
#define lson l,m,rt<<1
#define key_value ch[ch[root][1]][0]
#define rson m+1,r,rt<<1|1
#define pi acos(-1.0)
using namespace std;
const int MAXN = 200010;
int a[MAXN],n,ans[MAXN];
struct node{
    int len;
    int val;
    node(){}
    node(int tlen,int tval):len(tlen),val(tval){}
};
stack<node>s;
int main(){
    while(~scanf("%d",&n)){
        for(int i = 1; i <= n; i++){
            scanf("%d",&a[i]);
        }
        while(!s.empty())s.pop();
        memset(ans,0,sizeof(ans));
        for(int i = 1; i <= n; i++){
            int len = 0;//表示当前弹出的这些元素能够贡献多少的长度
            while(!s.empty()){
                node tp = s.top();
                if(tp.val < a[i])break;
                s.pop();
                int ret = tp.len + len;//这个元素能够贡献的长度 （也就是说这个元素已经统计过了多少长度，在这里也是可以连续的）
                ans[ret] = max(ans[ret],tp.val);
                len += tp.len;
            }
            s.push(node(len+1,a[i]));
        }
        int len = 0;
        while(!s.empty()){
            node tp = s.top();
            s.pop();
            int ret = len + tp.len;
            ans[ret] = max(ans[ret],tp.val);
            len += tp.len;
        }
        for(int i = n; i >= 1; i--){
            ans[i] = max(ans[i],ans[i+1]);
        }
        for(int i = 1; i <= n; i++){
            printf("%d ",ans[i]);
        }
        printf("\n");
    }
    return 0;
}