(单调栈)Bad Hair Day

Bad Hair Day

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 2   Accepted Submission(s) : 1

Problem 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

610374122

 

Sample Output

5

 

Source

PKU

 

思路分析:

使用"单调栈"。（从最后一头牛开始扫描（因为牛都往右看），并在最后的位置后面再虚拟一头hair无限高的牛，以方便处理。当处理完所有牛后便能得出最后答案）

痛的领悟：

答案要用long long类型，因为N最多取80000，而答案最多取N*（N+1）/2, 大于64亿。

<pre name="code" class="cpp">#include <cstdio>#include <algorithm>#include <math.h>#include <string.h>using namespace std;#define maxn 80005#define maxint 1000000005int h[maxn],s[maxn];int main(){int N, i;scanf("%d", &N);for (i = 0; i < N; i++)scanf("%d", &h[i]);int top = 0, base = 0;long long ans = 0;s[top++] = N;h[N] = maxint;//虚拟一头hair无限高的牛for (i = N - 1; i >= 0; i--){while (h[s[top - 1]] < h[i])//比当前牛的hair高度小的牛(编号)都出栈top--;ans += s[top - 1] - i-1;s[top++] = i;//只需要将牛的编号入栈便可得出答案}printf("%lld\n", ans);return 0;}