洛谷P2866 [USACO06NOV]糟糕的一天Bad Hair Day

Posted 2020-09-23 zbtrs

P2866 [USACO06NOV]糟糕的一天Bad Hair Day

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  • 153提交
• 题目提供者洛谷OnlineJudge
• 标签USACO2006云端
• 难度普及/提高-
• 时空限制1s / 128MB

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• 题目标题

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.

农民约翰的某N(1 < N < 80000)头奶牛正在过乱头发节！由于每头牛都意识到自己凌乱不堪 的发型，约翰希望统计出能够看到其他牛的头发的牛的数量.

每一头牛i有一个高度所有N头牛面向东方排成一排，牛N在最前面，而 牛1在最后面.第i头牛可以看到她前面的那些牛的头，只要那些牛的高度严格小于她的高度，而且 中间没有比hi高或相等的奶牛阻隔.

让N表示第i头牛可以看到发型的牛的数量；请输出Ci的总和

输入输出格式

输入格式：

 

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.

 

输出格式：

 

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

 

输入输出样例

输入样例#1：

6
10
3
7
4
12
2

输出样例#1：

5
分析：很显然是单调栈，至于单调栈的用法可以去看我的前几篇博客，这里计数器累加的top代表这只牛的头发能被前面的多少只牛看到（因为前面有top只牛比这只牛高）.

#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

int n,h[500010],stk[500010],top;
long long ans;

int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
        scanf("%d", &h[i]);

    for (int i = 1; i <= n; i++)
    {
        while (top != 0 && stk[top] <= h[i])
            top--;
        ans += top;
        stk[++top] = h[i];
    }
    printf("%lld", ans);

    return 0;
}