Codeforces Round #305 (Div. 1) B. Mike and Feet（单调栈）

B. 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

官方题解，学习单调栈了

For each i, find the largest j that aj < ai and
show it by li (if
there is no such j, then li = 0).

Also, find the smallest j that aj < ai and
show it by ri (if
there is no such j, then ri = n + 1).

This can be done in O(n) with a stack. Pseudo code of the first part (second part is also like that) :

stack s // initially empty
for i = 1 to n
while s is not empty and a[s.top()] >= a[i]
do s.pop()
if s is empty
then l[i] = 0
otherwise
l[i] = s.top()
s.push(i)

Consider that you are asked to print n integers, ans1, ans2, ..., ansn.
Obviously, ans1 ≥ ans2 ≥ ... ≥ ansn.

For each i, we know that ai can
be minimum element in groups of size 1, 2, ..., ri - li - 1.

Se we need a data structure for us to do this:

We have array ans1, ans2, ..., ansn and
all its elements are initially equal to 0. Also, n queries.
Each query gives x, val and want us to perform ans1 = max(ans1, val), ans2 = max(ans2, val), ..., ansx = max(ansx, val).
We want the final array.

This can be done in O(n) with a maximum partial sum (keeping maximum instead of sum), read
here for more information about partial sum.

Time complexity: 

.

C++ Code by PrinceOfPersia

C++ Code by Haghani

Java Code by Zlobober

#include <cstdio>
#include <iostream>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
typedef long long ll;
#define foreach(it,v) for(__typeof((v).begin()) it = (v).begin(); it != (v).end(); ++it)
const int inf = 0x3f3f3f3f;
const int maxn = 2e5 + 10;
int a[maxn],l[maxn],r[maxn],res[maxn];
int main(int argc, char const *argv[])
{
int n;
while(scanf("%d",&n)==1) {
for(int i = 1; i <= n; i++) scanf("%d",a+i);
stack<int>s;
for(int i = 1; i <= n; i++) {
res[i] = 0;
while(!s.empty()&&a[s.top()] >= a[i]) s.pop();
if(s.empty()) l[i] = 0;
else l[i] = s.top();
s.push(i);
}
while(!s.empty())s.pop();
for(int i = n; i >= 1; i--) {
while(!s.empty()&&a[s.top()] >= a[i]) s.pop();
if(s.empty()) r[i] = n + 1;
else r[i] = s.top();
s.push(i);
}
for(int i = 1; i <= n; i++) {
int x = r[i] - l[i] -1;
res[x] = max(res[x],a[i]);
}
for(int i = n-1; i > 0; i--) res[i] = max(res[i+1],res[i]);
for(int i = 1; i <= n; i++)printf("%d%c", res[i]," \n"[i==n]);
}
return 0;
}