CF305 Mike and Feet 單調棧（求每個長度對應的最小數字）

題目鏈接：http://codeforces.com/contest/548/problem/D

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 

單調棧：類似於poj2559的做法，將更新最大面積改爲更新ans。（可參照http://blog.csdn.net/alongela/article/details/8230739）

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int N = 200005;
struct node{
    int height,width;
};
node stack[N];//單調棧
int ans[N],h[N];//ans記錄結果，h輸入序列
int main()
{
	int n,x;
	while(~scanf("%d",&n)&&n){
        memset(ans,0,sizeof(ans));
	    for(int i=0;i<n;i++){
            scanf("%d",&x);
            h[i]=x;
	    }
	    h[n++]=0;//因爲輸入序列>0，所以後加0處理最後一個數就可以全部出棧
        int top=0;
        for(int i=0;i<n;i++){
            int tmp=0; //需要累加的寬度
            while(top>0&&stack[top-1].height>=h[i]){ 
                int ls=stack[top-1].width+tmp;
                if(stack[top-1].height>ans[ls]) ans[ls]=stack[top-1].height;
                tmp+=stack[top-1].width;
                --top;//出棧
            }
            stack[top].height=h[i];
            stack[top].width=tmp+1;
            ++top;//進棧
        }
        for(int i=n-1;i>=1;i--) ans[i]=max(ans[i+1],ans[i]);
        //單調棧只更新了值發生變化的點。所以在此處更新：未處理的點的值=（最近的寬度比它的點的值）
        
        for(int i=1;i<n-1;i++) printf("%d ",ans[i]);
        printf("%d\n",ans[n-1]);
	}
	return 0;
}