POJ 3250 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.

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

题意：牛i面向右看,当不比他高（包括等于）就是能看到，求这些牛能看到的和
单调栈的题目，保存每头牛能看到 最右端的坐标

代码如下：

#include<stdio.h> 
#include<string.h>
long long h[80005], right[80005];
int main()
{
    int i, j, k, n, temp;
    while(scanf("%d", &n)==1&&n)
    {
        memset(h, 0, sizeof(h));
        memset(right, 0, sizeof(right));
        for(i=1; i<=n; i++)
        {
            scanf("%lld", &h[i]);
            right[i]=i;
        }
        for(i=n-1; i>0; i--)
        {
            temp=i;
            while(h[temp+1]<h[i]&&temp<n) //想等也看不到 
                temp=right[temp+1];
            right[i]=temp;
        }
        long long area=0; //这里验证过了...long long 是可以AC的 
        for(i=1; i<=n; i++)
            area+=(right[i]-i);
        printf("%lld\n", area);
    }
}