Codeforces Round #305 (Div. 2) 548D Mike and Feet(单调栈)

D. Mike and Feet

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 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 

给定一个长度为n的序列，有n个起点，问你从每个起点出发相同长度的区间的最小值集合中的最大值。

单调栈性质：1.单调递增或递减 2.越靠近栈顶元素越后进栈。

与单调队列不同之处：只在栈顶操作，不限定大小。

进栈过程：对于单调递增栈，若当前进栈元素为e，从栈顶开始遍历元素，把小于e或者等于e的元素弹出栈，直接遇到一个大于e的元素或者栈为空为止，然后再把e压入栈中。对于单调递减栈，则每次弹出的是大于e或者等于e的元素。

AC代码：

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "stack"
using namespace std;
const int maxn = 2e5 + 5;
struct node
{
	/* data */
	int num, width;
	node() {}
	node(int n, int w): num(n),width(w) {}
};
stack s;
int a[maxn], ans[maxn];
int main(int argc, char const *argv[])
{
	int n;
	while(scanf("%d", &n) != EOF) {
		for(int i = 0; i < n; ++i)
			scanf("%d", &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;
}