Bad Hair Day
Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 21084   Accepted: 7202

Description

Some of Farmer John's N cows (1 ≤ N ≤ 80,000) are having a bad hair day! Since each cow is self-conscious about her messy hairstyle, FJ wants to count the number of other cows that can see the top of other cows' heads.

Each cow i has a specified height hi (1 ≤ h≤ 1,000,000,000) and is standing in a line of cows all facing east (to the right in our diagrams). Therefore, cow i can see the tops of the heads of cows in front of her (namely cows i+1, i+2, and so on), for as long as these cows are strictly shorter than cow i.

Consider this example:

        =
=       =
=   -   =         Cows facing right -->
=   =   =
= - = = =
= = = = = =
1 2 3 4 5 6

Cow#1 can see the hairstyle of cows #2, 3, 4
Cow#2 can see no cow's hairstyle
Cow#3 can see the hairstyle of cow #4
Cow#4 can see no cow's hairstyle
Cow#5 can see the hairstyle of cow 6
Cow#6 can see no cows at all!

Let ci denote the number of cows whose hairstyle is visible from cow i; please compute the sum of c1 through cN.For this example, the desired is answer 3 + 0 + 1 + 0 + 1 + 0 = 5.

Input

Line 1: The number of cows, N
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i.

Output

Line 1: A single integer that is the sum of c1 through cN.

Sample Input

6
10
3
7
4
12
2

Sample Output

5

题意:个子高的牛能看到个子低的牛,且每头牛的视线都是一个方向,问每头牛分别能看到视线方向上的多少头牛。
思路:单调栈,从后往前考虑队列,若当前的牛能看到存储在栈中的牛,则把栈中的牛从栈中抛出,直到栈为空或者栈中的牛的身高大于等于当前牛的身高为止,将当前的牛压入栈中,每次执行这样的操作即可。
AC代码:
#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<vector>
#include<cstring>
#include<string>
#include<cmath>
using namespace std;
#define INF 0x3f3f3f3f
#define N_MAX 80000+20
typedef long long ll;
int n,h[N_MAX];
int st[N_MAX],num[N_MAX];
int main() {
scanf("%d",&n);
for (int i = ; i < n; i++)scanf("%d",&h[i]);
int t = ;//栈的大小
for (int i = n-; i >=; i--) {
while (t > && h[st[t - ]] <h[i])t--;
num[i] = t == ? n - - i : st[t - ] - i - ;
st[t++]= i;
}
ll sum = ;
for (int i = ; i < n; i++)
sum += num[i];
printf("%lld\n",sum);
return ;
}
												

poj 3250 Bad Hair Day的更多相关文章

  1. Poj 3250 单调栈

    1.Poj 3250  Bad Hair Day 2.链接:http://poj.org/problem?id=3250 3.总结:单调栈 题意:n头牛,当i>j,j在i的右边并且i与j之间的所 ...

  2. poj 3250 Bad Hair Day (单调栈)

    http://poj.org/problem?id=3250 Bad Hair Day Time Limit: 2000MS   Memory Limit: 65536K Total Submissi ...

  3. poj 3250 Bad Hair Day(单调队列)

    题目链接:http://poj.org/problem?id=3250 思路分析:题目要求求每头牛看见的牛的数量之和,即求每头牛被看见的次数和:现在要求如何求出每头牛被看见的次数? 考虑到对于某头特定 ...

  4. (单调队列) Bad Hair Day -- POJ -- 3250

    http://poj.org/problem?id=3250 Bad Hair Day Time Limit: 2000MS   Memory Limit: 65536K Total Submissi ...

  5. poj 3250 Bad Hair Day(栈的运用)

    http://poj.org/problem?id=3250 Bad Hair Day Time Limit: 2000MS   Memory Limit: 65536K Total Submissi ...

  6. POJ 3250 Bad Hair Day(单调栈)

    [题目链接] http://poj.org/problem?id=3250 [题目大意] 有n头牛,每头牛都有一定的高度,他能看到在离他最近的比他高的牛前面的所有牛 现在每头牛往右看,问每头牛能看到的 ...

  7. POJ 3250 Bad Hair Day --单调栈(单调队列?)

    维护一个单调栈,保持从大到小的顺序,每次加入一个元素都将其推到尽可能栈底,知道碰到一个比他大的,然后res+=tail,说明这个cow的头可以被前面tail个cow看到.如果中间出现一个超级高的,自然 ...

  8. poj 3250 栈应用

    #include<iostream> #include<cstring> #include<algorithm> #include<cstdio> #d ...

  9. poj 3250 Bad Hair Day【栈】

    Bad Hair Day Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 15922   Accepted: 5374 Des ...

  10. poj 3250 Bad Hair Day 单调栈入门

    Bad Hair Day 题意:给n(n <= 800,000)头牛,每头牛都有一个高度h,每头牛都只能看到右边比它矮的牛的头发,将每头牛看到的牛的头发加起来为多少? 思路:每头要进栈的牛,将栈 ...

随机推荐

  1. ES6学习(二):函数的扩展

    chapter07 函数的扩展 7.1 函数默认值 7.1.1 参数默认值简介 传统做法的弊端(||):如果传入的参数相等于(==)false的话,仍会被设为默认值,需要多加入一个if判断,比较麻烦. ...

  2. Centos 安装python 3.7 ,同时兼容python2.7

    下载Python源码 从http://www.python.org/download/根据需要的版本下载源文件. 例如上图就是我在官网直接找到3.5.6版本的下载页面,点击的tar源码包进行下载. 1 ...

  3. Linux产生随机数的几种方法

    .echo $RANDOM .openssl rand -base64 .date +%n%N .head /dev/urandom |cksum .cat /proc/sys/kernel/rand ...

  4. crontab -e 和/etc/crontab的区别

    /etc/crontab文件和crontab -e命令区别/etc/crontab文件和crontab -e命令区别 1.格式不同 前者 # For details see man 4 crontab ...

  5. Python循环的一些基本练习

    #1:# name = input('请输入你的身份')# if name == 'egon':# print('--> 超级管理员')# elif name == 'tom':# print( ...

  6. thinkphp 3.2.3 - Think.class.php 解析

    class Think { public static function start() { // 注册AUTOLOAD方法 spl_autoload_register('Think\Think::a ...

  7. Python猫荐书系列之七:Python入门书籍有哪些?

    本文原创并首发于公众号[Python猫],未经授权,请勿转载. 原文地址:https://mp.weixin.qq.com/s/ArN-6mLPzPT8Zoq0Na_tsg 最近,猫哥的 Python ...

  8. C# datagridview列绑定类中类的属性

    原创作品,允许转载,转载时请务必以超链接形式标明文章 原始出处 .作者信息和本声明.否则将追究法律责任.http://www.cnblogs.com/linghaoxinpian/p/5906374. ...

  9. 云中Active Directory是如何工作的?

    [TechTarget中国原创] 微软公司1999年在Windows Server 2000中引入Active Directory功能.后期的Windows Server版本中陆续进行改善提升,Win ...

  10. Oracle数据库迁移--->从Windows到Linux

    I did a practice to migrate the oracle database from windows to linux operation system. The followin ...