POJ3250(单调栈)

Bad Hair Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 17614		Accepted: 5937

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

一群高度不完全相同的牛从左到右站成一排，每头牛只能看见它右边的比它矮的牛的发型，若遇到一头高度大于或等于它的牛，则无法继续看到这头牛后面的其他牛。

给出这些牛的高度，要求每头牛可以看到的牛的数量的和。

把要求作一下转换，其实就是要求每头牛被看到的次数之和。这个可以使用单调栈来解决。

从左到右依次读取当前牛的高度，从栈顶开始把高度小于或等于当前牛的高度的那些元素删除，此时栈中剩下的元素的数量就是可以看见当前牛的其他牛的数量。把这个数量加在一起，就可以得到最后的答案了。

单调栈的代码最后竟那么简洁。

 //2016.8.23
 #include<cstdio>

 const int N = ;
 int Stack[N], hi, top;

 int main()
 {
     int n;
     long long ans;
     while(scanf("%d", &n)!=EOF)
     {
         ans = top = ;
         for(int i = ; i <= n; i++)
         {
             scanf("%d", &hi);
             while(top>&&Stack[top-]<=hi)top--;
             ans += top;
             Stack[top++] = hi;
         }
         printf("%lld\n", ans);
     }

     return ;
 }