poj 3250 Bad Hair Day (单调栈)
http://poj.org/problem?id=3250
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 11473 | Accepted: 3871 |
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 ≤ hi ≤ 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
Output
Sample Input
6
10
3
7
4
12
2
Sample Output
5
Source
{
__int64 height;
__int64 startPos;
}node;
#include<iostream>
#include<stdio.h>
#include<stack> #define INF 1000000001 using namespace std; struct Nod
{
__int64 height;
__int64 startPos;
}node; int main()
{
__int64 n;
while(~scanf("%I64d",&n))
{
__int64 i,a,sum=;
stack <Nod> stk;
node.height = INF;
node.startPos = -;
stk.push(node);
for(i=;i<n;i++)
{
scanf("%I64d",&a);
while(a>=stk.top().height)
{
node = stk.top();
stk.pop();
sum += i - node.startPos - ;
}
node.height = a;
node.startPos = i;
stk.push(node);
}
while(!stk.empty())
{
node = stk.top();
stk.pop();
if(node.height!=INF) sum += i - node.startPos - ;
}
printf("%I64d\n",sum);
}
return ;
}
poj 3250 Bad Hair Day (单调栈)的更多相关文章
- 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 ...
- poj 3250 Bad Hair Day (单调栈)
Bad Hair Day Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 14883 Accepted: 4940 Des ...
- 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 ...
- poj 2769 感觉♂良好 (单调栈)
poj 2769 感觉♂良好 (单调栈) 比尔正在研发一种关于人类情感的新数学理论.他最近致力于研究一个日子的好坏,如何影响人们对某个时期的回忆. 比尔为人的一天赋予了一个正整数值. 比尔称这个值为当 ...
- POJ 2082 Terrible Sets(单调栈)
[题目链接] http://poj.org/problem?id=2082 [题目大意] 给出一些长方形下段对其后横向排列得到的图形,现在给你他们的高度, 求里面包含的最大长方形的面积 [题解] 我们 ...
- POJ 2796 Feel Good 【单调栈】
传送门:http://poj.org/problem?id=2796 题意:给你一串数字,需要你求出(某个子区间乘以这段区间中的最小值)所得到的最大值 例子: 6 3 1 6 4 5 2 当L=3,R ...
- POJ 2796 Feel Good(单调栈)
传送门 Description Bill is developing a new mathematical theory for human emotions. His recent investig ...
- Feel Good POJ - 2796 (前缀和+单调栈)(详解)
Bill is developing a new mathematical theory for human emotions. His recent investigations are dedic ...
- 51nod 1215 数组的宽度&poj 2796 Feel Good(单调栈)
单调栈求每个数在哪些区间是最值的经典操作. 把数一个一个丢进单调栈,弹出的时候[st[top-1]+1,i-1]这段区间就是弹出的数为最值的区间. poj2796 弹出的时候更新答案即可 #inclu ...
随机推荐
- highcharts 去掉右下角链接
去掉右下角的highcharts.com链接需要加入以下代码: credits: { enabled:false }, 如果不设置,那么默认为显示.
- oracle“记录被另一个用户锁住”
1.查看数据库锁,诊断锁的来源及类型: select object_id,session_id,locked_mode from v$locked_object; 或者用以下命令: select b. ...
- Memcached 工作原理
http://hzp.iteye.com/blog/1872664 Memcached处理的原子是每一个(key,value)对(以下简称kv对),key会通过一个hash算法转化成hash-key, ...
- 在DataTable中添加行和列数据
DataRow newRow = dtResult.NewRow(); newRow["ProName"] = "名字"; newRow["ProPr ...
- apache ambari web页面无法访问解决办法
ambari-server启动成功,但是页面无法访问 作者:Bo liang链接:http://www.zhihu.com/question/34405898/answer/115001510来源:知 ...
- MVC (M-V-C启动程序调用关系)
在网上有很多mvc程序启动,调用之间的关系与顺序.而且还有很多很不错的网站.推荐一个 http://www.cnblogs.com/QLeelulu/archive/2008/09/30/1 ...
- linux查看文件个数命令
linux下查看当前目录下文件个数命令: 使用背景:有时想了解一个目录下具体有多少文件或者有多少文件夹. 1. 查看当前目录下文件个数 ls -l |grep "^-"|wc -l ...
- OpenJudge 2811 熄灯问题 / Poj 1222 EXTENDED LIGHTS OUT
1.链接地址: http://bailian.openjudge.cn/practice/2811 http://poj.org/problem?id=1222 2.题目: 总时间限制: 1000ms ...
- 创建线程的两种方式比较Thread VS Runnable
1.首先来说说创建线程的两种方式 一种方式是继承Thread类,并重写run()方法 public class MyThread extends Thread{ @Override public vo ...
- BootstrapDialog.show函数底层简化
平台用的全部都是BootStrapDialog的弹窗,然后美工设计了一个统一的样式,每次写的时候,都要对其进行样式重写:写吐了快,所以对BootStrap.底层做了修改: 也就是说,只要你要写的界面包 ...