# Codeforces Round #305 (Div. 2)D---Mike and Feet(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.
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

``````/*************************************************************************
> File Name: D.cpp
> Author: ALex
> Mail: [email protected]
> Created Time: 2015年05月27日 星期三 15时45分40秒
************************************************************************/

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

static const int N = 200010;
int L[N], R[N];
PLL Stack[N];
int Top;
int arr[N];
PLL use[N];

int cmp(PLL a, PLL b) {
return a.first > b.first;
}

int main() {
int n;
while (~scanf("%d", &n)) {
for (int i = 1; i <= n; ++i) {
scanf("%d", &arr[i]);
use[i] = make_pair(arr[i], i);
}
Top = 0;
for (int i = n; i >= 1; --i) {
if (!Top) {
Stack[++Top] = make_pair(arr[i], i);
}
else {
while (Top) {
PLL u = Stack[Top];
if (u.first <= arr[i]) {
break;
}
--Top;
L[u.second] = i + 1;
}
Stack[++Top] = make_pair(arr[i], i);
}
}
while (Top) {
PLL u = Stack[Top];
--Top;
L[u.second] = 1;
}
for (int i = 1; i <= n; ++i) {
if (!Top) {
Stack[++Top] = make_pair(arr[i], i);
}
else {
while (Top) {
PLL u = Stack[Top];
if (u.first <= arr[i]) {
break;
}
--Top;
R[u.second] = i - 1;
}
Stack[++Top] = make_pair(arr[i], i);
}
}
while (Top) {
PLL u = Stack[Top];
--Top;
R[u.second] = n;
}
int len = 1;
sort(use + 1, use + 1 + n, cmp);
for (int i = 1; i <= n; ++i) {
int l = L[use[i].second];
int r = R[use[i].second];
for (int j = len; j <= r - l + 1; ++j) {
printf("%d ", use[i].first);
++len;
}
if (len > n) {
break;
}
}
printf("\n");
}
return 0;
}``````

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

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