D - Mike and Feet CodeForces - 548D（单调栈）

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.

Examples
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：1~n长度区间求最小值的最大值
思路：单调栈求解，留个坑，以后补吧。

#include <bits/stdc++.h>//炒鸡恶心的单调栈也是我的第一道单调栈

using namespace std;

const int N = 2e5 + 5;

struct node
{
    int num,width;
    node(){}
    node(int _num,int _width):num(_num),width(_width){}
};
stack<node>S;
int a[N],ans[N];

int main()
{
    int n;
    cin>>n;
    for(int i = 0;i < n;i++)
        cin>>a[i];
    a[n++] = 0;
    memset(ans,0,sizeof(ans));
    for(int i = 0;i < n;i++)
    {
        int len = 0;
        node k;
        while(!S.empty())
        {
            k = S.top();
            if(k.num < a[i])
            {
                break;
            }
            int ls = k.width + len;
            if(k.num > ans[ls])
            {
                ans[ls] = k.num;
            }
            len += k.width;
            S.pop();
        }
        S.push(node(a[i],len + 1));
    }
    for(int i = n - 1;i >= 1;i--)
    {
        ans[i] = max(ans[i],ans[i + 1]);
    }
    printf("%d",ans[1]);
    for(int i = 2;i < n;i++)
    {
        printf(" %d",ans[i]);
    }
    printf("\n");
    return 0;
}