SCUT J.O: Bad Hair Day

最新推荐文章于 2019-04-04 21:01:05 发布

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解决一个有趣的问题：计算每头牛能看到前方多少头牛的头顶，最终求得所有牛能看到的头顶总数。采用栈的数据结构实现高效算法。

Bad Hair Day

Time Limit:

2000MS

Memory Limit:

65536K

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 unsigned integer that is the sum of c1 through cN.

Sample Input

Sample Output

Hint

No hint.

Source

解体思路：应用栈，复杂度为θ(n)

#include <cstdio> typedef unsigned int base; base list[100000], sum = 0; int cnt, stack[100000], top = 0; int main(){ scanf("%d", &cnt); for (int i = 0; i < cnt; i ++) scanf("%u", list + i); list[cnt] = 4294967295; stack[top] = cnt; for (int i = cnt - 1; i > -1; i --){ while (list[i] > list[stack[top]]) top --; sum += stack[top] - i - 1; stack[++ top] = i; } printf("%u/n", sum); return 0; }