poj 3250 Bad Hair Day

Bad Hair Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10031		Accepted: 3326

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

#include<iostream>
using namespace std;
int h[80010];
int r[80010];
__int64 mmax;

void cal(int m)
{
    memset(r, 0, sizeof(r));
    mmax = 0;
    int i = 0;
    
    for(i = 1; i <= m; ++i)
    {
        r[i] = i;
    }
    for(i = m; i >= 1; --i)
    {
        while(h[r[i]+1] < h[i])
            r[i] = r[r[i]+1];
    }
    for(i = 1; i <= m; ++i)
    {
        mmax += (r[i] - i);
    }
}

int main()
{
    int n;
    int i;
//    freopen("E:\\c++\\oj\\t.txt", "r", stdin);
//    freopen("E:\\c++\\oj\\out.txt", "w", stdout);
    while(cin>>n)
    {
        for(i = 1; i <= n; ++i)
        {
            scanf("%d",&h[i]);
        }
        h[0] = h[n+1] = INT_MAX;
        cal(n);
    }
    cout<<mmax<<endl;
}