【POJ - 3250】【 Bad Hair Day 】

题目：

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

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 c 1 through cN.

Sample Input

6
10
3
7
4
12
2

Sample Output

5

题意：

jhon对自己的头发不是很满意，然后就去看别人的发型，然后奶牛们排成一排，且面朝一个方向，问每个奶牛能看到奶牛数目的和是多少。。。

解题思路：

单调栈的入门题，转变一下思路，从每个顶点能看到多少脑袋，转换为能被多少人看到，且看到的情况是每个人的后方有多少比它高的存在（且存在遮挡现象，所以会pop）

ac代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<stack>
#include<queue>
#define maxn 801000
using namespace std;
stack <long long int >q;
int main()
{
	int n;
	while(q.size()) q.pop();
	while(scanf("%d",&n)!=EOF)
	{
		long long int sum=0;
		long long int x;
		cin>>x;
		q.push(x);
		for(int i=1;i<n;i++)
		{
			cin>>x;
			while(!q.empty()&&q.top()<=x)
				q.pop();
			sum+=q.size();
			q.push(x);
		}
		cout<<sum<<endl;
	}
	return 0;

}