（单调队列） Bad Hair Day -- POJ -- 3250

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

Bad Hair Day Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 15956   Accepted: 5391 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 hi (1 ≤ hi ≤ 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 ci denote the number of cows whose hairstyle is visible from cow i; please compute the sum of c1 through cN.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 cN. Sample Input 6 10 3 7 4 12 2 Sample Output 5 Source USACO 2006 November Silver

#include<stdio.h>
#define N 80010

int Stack[N];

int main()
{
    int top = , i, n, p;
    __int64 sum = ;

    scanf("%d", &n);

    for(i=; i<=n; i++)
    {
        scanf("%d", &p);
        while(top> && Stack[top]<=p)
            top--;
        sum += top;
        Stack[++top] = p;
    }

    printf("%I64d\n", sum);
    return ;
}