POJ 3250 Bad Hair Day

Bad Hair Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 11337		Accepted: 3831

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
代码：

#include<cstdio>
#include<stack>
#include<iostream>
using namespace std ;
#define MAXN 80002
stack<int>s ;
int main()
{
    int i , j , n , m ;
    long long  sum ;
    while( cin >> n )
    {
        cin >> j ;
        sum = 0 ;
        while( !s.empty() ) 
            s.pop() ;
        s.push( j ) ;
        // 建立一个单调递增栈
        for( i = 1; i < n ;i++)
        {
            scanf( "%d" , &j ) ;
            while( !s.empty() && j >= s.top() ) s.pop() ;
            sum += s.size() ;
            // 左边有多少个比当前高 ，栈里面就有多少个元素
            s.push( j ) ;
        }
        cout << sum << endl;
    }
}