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

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

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.

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.

Print n integers in one line. For each x from 1 to n, print the maximum strength among all groups of size x.

10
1 2 3 4 5 4 3 2 1 6

6 4 4 3 3 2 2 1 1 1

這題可以用單調棧做，維護一個棧，記錄minmum(該區間的最小值）和count（區間的總長度）。

#include
#include
#include
#include
#include
using namespace std;
#define maxn 200060
int ans[maxn];
struct node{
	int count,minmum;
}stack[maxn];
int main()
{
	int n,m,i,j,top,count,b;
	while(scanf(%d,&n)!=EOF)
	{
		memset(ans,0,sizeof(ans));
		top=0;
		for(i=1;i<=n;i++){
			scanf(%d,&b);
			count=0;
			while(top>0 && stack[top].minmum>=b){
				stack[top].count+=count;
				count=stack[top].count;
				if(ans[count]0){
				stack[top].count+=count;
				count=stack[top].count;
				if(ans[count]
                /*這裡算出來的ans[i]是連續長度為i的區間的最小值，但這個最小值是所有連續長度為i的區間長度的最大值，下面如果ans[i+1]比ans[i]大，那麼ans[i]可以更新為ans[i+1],因為如果i+1個連續數區間的最小值的最大值是b,那麼去掉一個數，一定可以做到長度為i的連續數區間的最大值是b。*/
		for(i=n;i>=2;i--){
			if(ans[i]>ans[i-1]){
				ans[i-1]=ans[i];
			}
		}
		for(i=1;i<=n-1;i++){
			printf(%d ,ans[i]);
		}
		printf(%d
,ans[i]);
	}
	return 0;
}