poj3250 Bad Hair Day (单调栈)

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.

单调栈裸题

#include 
#include 
#include 
#include 
using namespace std;

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        stack  sta;
        int t;
        long long sum=0;
        scanf("%d",&t);
        sta.push(t);
        n--;
        while(n--)
        {
            scanf("%d",&t);
            while(sta.size()!=0&&sta.top()<=t) sta.pop();
            sum+=sta.size();
            sta.push(t);
        }
        printf("%lld\n",sum);
    }
    return 0;
}