poj 3250 Bad Hair Day 【单调栈】

Bad Hair Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15953		Accepted: 5388

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 integer that is the sum of c1 through cN.
Sample Input

6
10
3
7
4
12
2

Sample Output
5

题意：有一排牛都向右看，只有左边的牛比右边的牛高才能看到右边的牛的头顶，问从第一头牛开始到n头牛为止能看到的牛的头顶的总和。

思路： 表示根本就没想到用单调栈，每加入一个新元素，就将栈中的小于等于其的元素弹出。因为如果加入的新元素大于或等于栈中的元素，那么栈中的元素是看不到新元素的头顶的。每次把弹出后栈中的剩余元素的个数相加，在将新元素压入栈中。相加的总和就是总个数。

<pre name="code" class="cpp">#include<stdio.h>
#include<stack>
using namespace std;
int main()
{
int n,i;
int a,b;
long long sum;
stack<int>s;
while(~scanf("%d",&n))
{

scanf("%d",&a);
s.push(a);
sum=0;
for(i=1;i<n;i++)
{
scanf("%d",&b);
while(!s.empty()&&b>=s.top())
{
s.pop();
}
sum+=s.size();
s.push(b);
}
printf("%lld\n",sum);
}
return 0;
}