Bad Hair Day(单调栈)

Bad Hair Day

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:

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

这里是引用
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

• 若准备进去的元素大于或等于栈顶元素,则删除栈顶元素,继续与下一个栈中元素做比较直到没有元素(或者小于栈顶元素),然后将其入栈
• cnt表示如果能看到i,那么就相当于看到了i之前的cnt个(包括i)

#include<iostream>
#include<algorithm>
#include<stack>
using namespace std;
typedef unsigned long long ll;int main()
{
    int N;ll cnt=0;cin>>N;stack<int> stk;for(int i=0;i<N;i++){
    int x;cin>>x;while(!stk.empty() && stk.top()<=x){
    stk.pop();}cnt+=stk.size();stk.push(x);}cout<<cnt<<endl;return 0;
}