codeforces 548 D. Mike and Feet (單調棧)

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 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 

思路：
單調棧預處理，對於每一個數，以它爲最小值的最左的左下標L，最右的右下標R，然後更新答案ans[R-L+1]就好了，可以用單調棧搞定

/*======================================================
# Author: whai
# Last modified: 2015-12-09 13:39
# Filename: d.cpp
======================================================*/
#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <stack>

using namespace std;

#define LL __int64
#define PB push_back
#define P pair<int, int>
#define X first
#define Y second

const int N = 2 * 1e5 + 5;
int a[N];
int L[N], R[N];

P st[N];
int top = -1;

int ans[N];

void gao(int n) {
	top = -1;
	st[++top] = P(-1, 0);
	for(int i = 1; i <= n; ++i) {
		int j;
		for(j = top; j >= 0; --j) {
			if(a[i] >= st[j].X) break;
			else R[st[j].Y] = i - 1;
		}
		if(a[i] > st[j].X) L[i] = st[j].Y + 1;
		else L[i] = L[st[j].Y];
		top = j;
		st[++top] = P(a[i], i);
	}
	for(int i = 0; i <= top; ++i) {
		R[st[i].Y] = n;
	}
	for(int i = 1; i <= n; ++i) {
		//cout<<L[i]<<' '<<R[i]<<endl;
		int u = R[i] - L[i] + 1;
		ans[u] = max(ans[u], a[i]);
	}
	for(int i = n - 1; i > 0; --i) {
		ans[i] = max(ans[i], ans[i + 1]);
	}
	for(int i = 1; i <= n; ++i) {
		printf("%d ", ans[i]);
	}
}

int main() {
	int n;
	while(cin>>n) {
		for(int i = 1; i <= n; ++i) {
			scanf("%d", &a[i]);
		}
		gao(n);
	}
	return 0;
}