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Bad Hair Day 单调栈

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Bad Hair Day
Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 13562   Accepted: 4576

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

Source

USACO 2006 November Silver

题目大意:从左往右,每个牛都有一个高度,他能看到右边低于他高度的牛,问总共能看到多少,,

反着想,看这个牛能被多少牛看到,要是左边的有比他低或等于他就不被看到,直接去掉就好,

ac代码

 

[cpp] view plain copy
 
 print?
  1. #include<stdio.h>  
  2. #include<stack>  
  3. #include<string>  
  4. #include<iostream>  
  5. using namespace std;  
  6. int main()  
  7. {  
  8.     int n;  
  9.     while(scanf("%d",&n)!=EOF)  
  10.     {  
  11.         int a[100010];  
  12.         int i;  
  13.         __int64 ans=0;  
  14.         stack<int>s;  
  15.         for(i=0;i<n;i++)  
  16.         {  
  17.             scanf("%d",&a[i]);  
  18.         }  
  19.         for(i=0;i<n;i++)  
  20.         {  
  21.             while(!s.empty()&&s.top()<=a[i])  
  22.                 s.pop();  
  23.             ans+=s.size();  
  24.             s.push(a[i]);  
  25.         }  
  26.         printf("%I64d\n",ans);  
  27.     }  
  28. }  

Bad Hair Day 单调栈

原文:http://www.cnblogs.com/yzm10/p/7197507.html

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