POJ 3250 Bad Hair Day (栈)

Bad Hair Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15939		Accepted: 5381

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 c ₁ through c_N.

Sample Input

6
10
3
7
4
12
2

Sample Output

5

题意：给出n个牛的高度，每个牛可以看到他后面的比他矮的且没有被挡住得求，问左右牛看到牛的个数和
思路：因为n比较大，可以换种想法，求每个牛可以被他前边几头牛看到，前边比他高的可以看到他，设个
单调栈，他比栈顶大就一直退栈就好了，最后栈的size就是它可以被前边的牛看到的数目，然后他进栈，把
每一位的可行都加起来就是最后的ans了，注意为longlong
ac代码：
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stack>
#include<iostream>
#include<algorithm>
#define fab(a) (a)>0?(a):(-a)
#define LL long long
#define MAXN 100100
#define INF 0xfffffff 
using namespace std;
int main()
{
	int n,i,j;
	LL a;
	while(scanf("%d",&n)!=EOF)
	{
		stack<LL>s;
		LL ans=0;
		scanf("%lld",&a);
		s.push(a);
		for(i=1;i<n;i++)
		{
			scanf("%lld",&a);
			while(!s.empty()&&s.top()<=a)
			s.pop();
			//printf("%d\n",s.top());
			ans+=s.size();
			s.push(a);
		}
		printf("%lld\n",ans);
	}
	return 0;
}