POJ——T 3250 Bad Hair Day

http://poj.org/problem?id=3250

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 19619		Accepted: 6702

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

Sample Output

Source

USACO 2006 November Silver

 #include <algorithm>

 #include <iostream>

 #include <cstdio>

 using namespace std;

 #define LL long long

 const int N();

 LL Stack[N],top,ans;

 int main()

 {

     LL n; cin>>n;

     for(LL h,i=;i<=n;i++)

     {

         cin>>h;

         for(;top&&Stack[top]<=h;) top--;

         ans+=top; Stack[++top]=h;

     }

     cout<<ans;

     return ;

 }

巴特西

POJ——T 3250 Bad Hair Day

http://poj.org/problem?id=3250

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