【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 ≤ h≤ 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

分析:单调栈模板题!

注意:题目上说是严格单调,也就是说栈内没有重复的,也就是说弹出时是大于等于号,因为这玩意错了五次自闭了。。。

#include <iostream>
#include <cmath>
#include <algorithm>
#include <queue>
#include <functional>
#include <cstdio>
#include <stack>
using namespace std;
typedef long long LL;
int main()
{
  int n = 0;
  while (cin >> n)
  {
    LL sum = 0;
    stack<LL> sta;
    for (int i = 0; i < n; i++)
    {
      LL temp;
      scanf("%lld", &temp);
      while (!sta.empty() && temp >= sta.top()) sta.pop();//单调栈 栈底为大
      sum += sta.size();
      sta.push(temp);
    }
    cout << sum << endl;
  }
}

 

 

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