Mike and Feet(单调栈)

Description
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 Input
Input
10
1 2 3 4 5 4 3 2 1 6
Output
6 4 4 3 3 2 2 1 1 1

就是要我们在i的长度里面找出最小的，然后每个i相等长度的又找出最大的来；

#include 
#include 
#include 
#include 
#include 
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;
}