POJ 3250 Bad Hair Day【单调栈】

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 integer that is the sum of c 1 through cN.
Sample Input
6
10
3
7
4
12
2
Sample Output
5325

#include<iostream>#include<string>#include<algorithm>#include<functional>using namespace std;#define Irish_Moonshine main#define maxn 1000000#define LL long long intLL stack[maxn];LL ans, n, tmp,top;int Irish_Moonshine(){    while (~scanf("%d", &n))    {        top = 0; ans = 0;        for (int i = 0; i < n; i++)        {            scanf("%lld", &tmp);            while (tmp >= stack[top]&&top>=1)            {                top--;            }            ans += top;            stack[++top] = tmp;        }        printf("%lld\n", ans);    }    return 0;}