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
大概思路

#include <stdio.h>
#include <stack>
#include <algorithm>
using namespace std;
int main()
{
	int n;
	long long c=0;
	scanf("%d",&n);
	stack<int> s;  //栈存放的是下标
	int h[n+5],sum[n+5];
	for(int i=0;i<n;i++)
		scanf("%d",&h[i]);
	for(int i=0;i<n;i++)
	{
		if(s.empty()||h[s.top()]>h[i])
		    s.push(i);
		else
		{
			sum[s.top()]=i-s.top()-1;  //判断准备出栈的牛和准备进栈的牛之间有几头牛，-1减的是自己，因为下标从0开始，所以下标=前面有几头牛
			s.pop();
			i--;
		}
	}
	while(!s.empty())
	{
		sum[s.top()]=n-s.top()-1;  //此时栈中的牛右边的牛都比他矮，所以是总头数减去他的下标再减去他自己
		s.pop();
	}
	for(int i=0;i<n;i++)
	    c+=sum[i];
	printf("%lld\n",c);
	return 0;
}