Bad Hair Day 单调栈

Posted 2020-09-28 博客就叫Molex好了

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 h_i (1 ≤ h_i ≤ 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 c_i denote the number of cows whose hairstyle is visible from cow i; please compute the sum of c₁ through c_N.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 c_N.

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. }