Bad Hair Day（单调栈）

题目链接

题目描述：

Some of Farmer John's N cows (1 ≤ N ≤ 80,0001≤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 ≤ h_i ≤ 1,000,000,0001≤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.

Example：

input：

6
10
3
7
4
12
2

output：

5

题意：一群身高不等的奶牛排成一排，向右看，每个奶牛只能看到身高小于自己的奶牛发型，问这些奶牛能够看到的发型总和是多少
思路：利用单调栈（栈中元素从栈顶往下越来越大）的思想，可以计算出，每只奶牛能被他前面的多少只奶牛看到自己的发型，就反向得到了奶牛看到的发型总和
借用别人的解释：首先弹出元素是因为它右边相邻牛比它高（看不到它的头发并且挡住了该牛的视线），“挡住”这个关键字很重要，这就是说该牛已经看不到后面的其他牛的头发啦，而思路一是按顺序比较身高，统计每头牛前面能看到它头发的牛数，既然该牛望不到后面，那么把它出栈对整个结果没有影响。
例如：10 3 7 4 12 2

①Height_List：10 入栈首身高
②3<10，不弹出，num=0+1（当前栈中元素数），3入栈后 Height_List：10 3
③7 > 3（3 弹出） 7<10（10保留），栈中剩 10 ，num=1+1，7入栈后 Height_List：10 7
④4 < 7，没有弹出，栈中10 7都能看到4，num=2+2 ，4入栈后 Height_List：10 7 4
⑤12大于栈中全部元素，表示栈中所有元素看不到12，全部出栈，num不变，12入栈后 Height_List：12
⑥2 < 12，没有弹出，12能看到2，num=4+1
最后结果num=5

#include<bits/stdc++.h>
#include<iostream>
#include<cctype>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stack>
using namespace std;
typedef unsigned long long ull;
typedef  long long ll;
typedef pair<ll,ll> pi;
#define IOS std::ios::sync_with_stdio(false)
#define ls p<<1
#define rs p<<1|1
#define mod 1000000000 + 7
#define PI acos(-1.0)
#define INF   1e9
#define N 200000 + 5
/*********************Code*********************/
int n,a[N],ans;
stack<int>st;
int main(void){
    IOS;
    cin>>n;
    for(int i = 0;i <n;i++) //如果向左看，逆序输入
        cin>>a[i];
        for(int i = 0;i < n;i++){
            while(!st.empty()&&st.top()<=a[i])
                st.pop();
            ans +=st.size();
            st.push(a[i]);
        }
        cout<<ans<<endl;
        return 0;
}