POJ 3250 Bad Hair Day

Bad Hair Day

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

//思路转换成求每个牛被看到的次数之和，于是可以用一个单调栈来维护，依次读入数据，删除栈中所有小于当前牛高度的牛，剩下的就是可以看到当前牛发型的牛了，再把当前牛存进栈里
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
#include <cctype>
#include <vector>
#include <cmath>
using namespace std;

int main()
{
    int n, m;
    int s[80005];

    while(~ scanf("%d", &n))
    {
        int top = 0;
        long long res = 0;
        for(int i = 0; i < n; i++)
        {
            scanf("%d", &m);
            while(top > 0 && s[top-1] <= m) top--;
            res += top;
            s[top++] = m;
        }
        printf("%lld\n", res);
    }

	return 0;
}