单调栈之Bad Hair Day题解

题目：

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

5
题意：

每个奶牛都有一个高度，奶牛们排成一排，且面朝右方向，问每个奶牛能看到奶牛数目的和是多少

• 解题思路：

可以使用单调栈来解决此题。从每只奶牛能看到多少脑袋，转换为它能被多少人看到，即每只奶牛后方(注意：奶牛是面朝右边的)有多少比它高的存在（存在遮挡现象，如果某只奶牛后方的一只或者多只奶牛比它矮，那么这一只或者多只就需要出栈）

该题源自于北大OJ：3250 -- Bad Hair Day 

具体的解题思路如下，以该题的样例为例：10 3 7 4 12 2

1、将第1个元素入栈

2、将第2个元素与栈顶元素进行比较

        如果第2个元素大于栈顶元素就将栈顶元素出栈，代表此时第1个元素看不到第2个元素。

        如果第2个元素小于等于栈顶元素。此时计数变量+1，代表第1个元素可以看到第2个元素，并且将第2个元素入栈。 将第3个元素与栈顶元素进行比较

3、将第3个元素与栈顶元素进行比较

        如果第3个元素大于栈顶元素就将栈顶元素出栈，代表此时第2个元素看不到第3个元素，出栈后，让第3个元素再次与栈顶元素比较，直到栈空或者第3个元素小于等于栈顶元素。

        如果第3个元素小于等于栈顶元素。此时计数变量+栈中的元素个数，代表栈中的每个元素都可以看到第3个元素，之后以此类推，如下图演示

AC代码：

注意：这里要注意数据的范围，计数变量sum可能会超过int范围，需要定义为long long类型

#include<iostream>
#include<stack>
using namespace std;
int a[100000];
long long  sum=0;
stack<int> s;
//6 10 3 7 4 12 2
void nextLesser(int a[],int n)
{for(int i=1;i<=n;i++){//单调栈，>= while(!s.empty()&&s.top()<=a[i]){s.pop();	}sum+=s.size();s.push(a[i]); }	
}
int main()
{int n;scanf("%d",&n); for(int i=1;i<=n;i++){scanf("%d",&a[i]);}nextLesser(a,n);cout<<sum;return 0;
}

总结： 

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