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  • 洛谷——P2866 [USACO06NOV]糟糕的一天Bad Hair Day

    https://www.luogu.org/problem/show?pid=2866

    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

     1 #include <algorithm>
     2 #include <iostream>
     3 #define maxn 1000005
     4 
     5 using namespace std;
     6 
     7 long long n,ans,top,x;
     8 long long stack[maxn];
     9 
    10 int main()
    11 {
    12     cin>>n;
    13     for(int i=1;i<=n;i++)
    14     {
    15         cin>>x;
    16         while(top&&stack[top]<x)
    17             top--;
    18         ans+=top;
    19         stack[++top]=x;
    20     }
    21     cout<<ans;
    22     return 0;
    23 }
    熟悉栈
    ——每当你想要放弃的时候,就想想是为了什么才一路坚持到现在。
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  • 原文地址:https://www.cnblogs.com/Shy-key/p/6572202.html
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