NC25084 [USACO 2006 Nov S]Bad Hair Day

题目

题目描述

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.

输入描述

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.

输出描述

Line 1: A single integer that is the sum of c1 through cN.

示例1

输入

6
10
3
7
4
12
2

输出

5

题解

知识点：单调栈。

题目要求，每个牛的右侧的连续区间里身高严格比它小的牛的数量。转换思路，不去找小的牛的数量，而找到右侧第一个身高大于等于它的牛的位置即可，然后端点相减即要求数量，最后对每个牛的结果累加即可，用单调栈从右到左维护。

时间复杂度 \(O(n)\)

空间复杂度 \(O(n)\)

代码

#include <bits/stdc++.h>

using namespace std;

int h[80007], r[80007];

int main() {
    std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int n;
    cin >> n;
    for (int i = 0;i < n;i++) cin >> h[i];
    stack<int> s;
    for (int i = n - 1;i >= 0;i--) {
        while (!s.empty() && h[s.top()] < h[i]) s.pop();///因为严格小于才能看到，所以等于的身高不能弹出
        r[i] = s.empty() ? n - 1 : s.top() - 1;
        s.push(i);
    }
    long long sum = 0;
    for (int i = 0;i < n;i++) sum += r[i] - i;
    cout << sum << '\n';
    return 0;
}