poj 3250 Bad Hair Day【栈】

Bad Hair Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15922		Accepted: 5374

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

题意：每头牛都有身高，他（a）前边站的也有牛，有比他高的有比他低的，他能看见他前边比他低的，但是当有比他高的牛（b）的时候，这个高牛b前边的比a牛低的  a牛就看不见了（符合日常生活，当前边有一个比自己高的牛时，自己的视线就被挡住了）

题解：每次让牛入栈，当入栈的牛比栈顶的牛低时，继续入栈，直到遇到比栈顶高的牛弹出栈顶，

#include<stdio.h>
#include<stack>
#include<iostream>
#include<string.h>
using namespace std;
stack<int>s;
int cow[80005];
int main()
{
	int n,i;
	int sum=0;
	scanf("%d",&n);
	for(i=1;i<=n;i++)
		scanf("%d",&cow[i]);
	s.push(cow[1]);
	for(i=2;i<=n;i++)
	{
		while(!s.empty()&&s.top()<=cow[i])
		    s.pop();
		sum+=s.size();
		s.push(cow[i]);
	}
	printf("%u\n",sum);
	return 0;
}