Bad Hair Day（单调栈 ）

Bad Hair Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15941		Accepted: 5382

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
代码：

#include<iostream>
#include<cstdio>
#include<stack>
#include<algorithm>
using namespace std;
typedef long long LL;
int main(){
int n;
while(~scanf("%d",&n)){
stack<int>S;
int t;
scanf("%d",&t);
S.push(t);
LL ans=0;
for(int i=1;i<n;i++){
scanf("%d",&t);
while(!S.empty()&&t>=S.top())S.pop();
ans+=S.size();
S.push(t);
}
printf("%lld\n",ans);
}
return 0;
}