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  c ₁ 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：

View Code

#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);

    }

}