Codeforces Round #305 (Div. 2) D. Mike and Feet 单调栈
1 second
256 megabytes
standard input
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.
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
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x) cout<<"bug"<<x<<endl;
const int N=2e5+,M=1e6+,inf=;
const ll INF=1e18+,mod=1e9+;
/// 数组大小
int a[N],ans[N];
int l[N],r[N],s[N];
int main()
{
int n;
scanf("%d",&n);
for(int i=;i<=n;i++)
scanf("%d",&a[i]);
int st=;
s[++st]=;
for(int i=;i<=n;i++)
{
while(a[s[st]]>=a[i])st--;
l[i]=s[st];
s[++st]=i;
}
st=;
s[++st]=n+;
for(int i=n;i>=;i--)
{
while(a[s[st]]>=a[i])st--;
r[i]=s[st];
s[++st]=i;
}
for(int i=;i<=n;i++)
{
int len=r[i]-l[i]-;
ans[len]=max(ans[len],a[i]);
}
for(int i=n;i>=;i--)
ans[i]=max(ans[i],ans[i+]);
for(int i=;i<=n;i++)
printf("%d ",ans[i]);
return ;
}
Codeforces Round #305 (Div. 2) D. Mike and Feet 单调栈的更多相关文章
- Codeforces Round #305 (Div. 1) B. Mike and Feet 单调栈
B. Mike and Feet Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/547/pro ...
- set+线段树 Codeforces Round #305 (Div. 2) D. Mike and Feet
题目传送门 /* 题意:对于长度为x的子序列,每个序列存放为最小值,输出长度为x的子序列的最大值 set+线段树:线段树每个结点存放长度为rt的最大值,更新:先升序排序,逐个添加到set中 查找左右相 ...
- Codeforces Round #305 (Div. 2) D. Mike and Feet
D. Mike and Feet time limit per test 1 second memory limit per test 256 megabytes input standard inp ...
- Codeforces Round #305 (Div. 2)D. Mike and Feet(单调栈)
题意 n个值代表n个熊的高度 对于size为x的group strength值为这个group(连续的几个熊)中熊的最小的height值 对于x(1<=x<=n) 求出最大的strengt ...
- 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 besid ...
- 数论/暴力 Codeforces Round #305 (Div. 2) C. Mike and Frog
题目传送门 /* 数论/暴力:找出第一次到a1,a2的次数,再找到完整周期p1,p2,然后以2*m为范围 t1,t2为各自起点开始“赛跑”,谁落后谁加一个周期,等到t1 == t2结束 详细解释:ht ...
- 暴力 Codeforces Round #305 (Div. 2) B. Mike and Fun
题目传送门 /* 暴力:每次更新该行的num[],然后暴力找出最优解就可以了:) */ #include <cstdio> #include <cstring> #includ ...
- 字符串处理 Codeforces Round #305 (Div. 2) A. Mike and Fax
题目传送门 /* 字符串处理:回文串是串联的,一个一个判断 */ #include <cstdio> #include <cstring> #include <iostr ...
- Codeforces Round #305 (Div. 2) B. Mike and Fun 暴力
B. Mike and Fun Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/548/pro ...
随机推荐
- Docker 后台进程参数-------更改Docker运行根目录的方法
参数 介绍 --api-enable-cors=false 远程API调用. -b, --bridge="" 桥接一个系统上的网桥设备到 Docker 容器里,当使用 none 可 ...
- 使用pssh进行并行批量操作
假如同时给上千台服务器执行一个命令,拷贝一个文件,杀一个进程等,有什么简化运维管理的工具呢?在小型使用中我都是使用for循 环,数量巨大,一方面不确定操作是否成功,一方面for循环语句性能不好估计且是 ...
- BFS广搜题目(转载)
BFS广搜题目有时间一个个做下来 2009-12-29 15:09 1574人阅读 评论(1) 收藏 举报 图形graphc优化存储游戏 有时间要去做做这些题目,所以从他人空间copy过来了,谢谢那位 ...
- ComBSTR的使用
用 CComBSTR 进行编程 Visual Studio .NET 2003 3(共 3)对本文的评价是有帮助 - 评价此主题 ATL 类 CComBSTR 提供对 BSTR 数据类型的包装 ...
- 【weka】分类,cross-validation,数据
一.分类classifier 如何利用weka里的类对数据集进行分类,要对数据集进行分类,第一步要指定数据集中哪一列做为类别,如果这一步忘记了(事实上经常会忘记)会出现“Class index is ...
- python 多线程小练习
需求:有100个数据,启动5个线程,每个线程分20个数据,怎么把这20个数据分别传给每个线程. 1. 利用多线程实现 import threading nums = list(range(100)) ...
- jmeter 逻辑控制器Logic Controller详解
Jmeter之逻辑控制器(Logic Controller) 前言: 1. Jmeter官网对逻辑控制器的解释是:“Logic Controllers determine the order in w ...
- C语言typeof详解
前言: typeof关键字是C语言中的一个新扩展,这个特性在linux内核中应用非常广泛. 一,说明 typeof的参数可以是两种形式:表达式或类型. 1,表达式的的例子: ...
- windows脚本的if语句
- 模仿WIN32程序处理消息
#include "stdafx.h" #include "MyMessage.h" #include <conio.h> using namesp ...