522A. Reposts

最新推荐文章于 2020-03-14 20:14:23 发布

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最新推荐文章于 2020-03-14 20:14:23 发布

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分类专栏：算法和数据结构

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本文概述了AI音视频处理领域的关键技术，包括视频分割、语义识别、自动驾驶、AR、SLAM等，旨在为读者提供对AI音视频处理领域的全面理解。

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One day Polycarp published a funny picture in a social network making a poll about the color of his handle. Many of his friends started reposting Polycarp's joke to their news feed. Some of them reposted the reposts and so on.

These events are given as a sequence of strings "name1 reposted name2", where name1 is the name of the person who reposted the joke, and name2 is the name of the person from whose news feed the joke was reposted. It is guaranteed that for each string "name1 reposted name2" user "name1" didn't have the joke in his feed yet, and "name2" already had it in his feed by the moment of repost. Polycarp was registered as "Polycarp" and initially the joke was only in his feed.

Polycarp measures the popularity of the joke as the length of the largest repost chain. Print the popularity of Polycarp's joke.

Input

The first line of the input contains integer n (1 ≤ n ≤ 200) — the number of reposts. Next follow the reposts in the order they were made. Each of them is written on a single line and looks as "name1 reposted name2". All the names in the input consist of lowercase or uppercase English letters and/or digits and have lengths from 2 to 24 characters, inclusive.

We know that the user names are case-insensitive, that is, two names that only differ in the letter case correspond to the same social network user.

Output

Print a single integer — the maximum length of a repost chain.

Sample test(s)

Input

5
tourist reposted Polycarp
Petr reposted Tourist
WJMZBMR reposted Petr
sdya reposted wjmzbmr
vepifanov reposted sdya

Output

Input

6
Mike reposted Polycarp
Max reposted Polycarp
EveryOne reposted Polycarp
111 reposted Polycarp
VkCup reposted Polycarp
Codeforces reposted Polycarp

Output

Input

1
SoMeStRaNgEgUe reposted PoLyCaRp

Output

题意：输入一个n。输入一个人名a，reposted 和另一个人名b，n行，第一行的人名b一定是polycarp（不区分大小写）问人名传递的这个从polycarp开始的链的总长度。

思路：map一个容器s map<string,int > s 然后让字符串和字符串的出现时链长对应起来，后面每输入一组人名，就让前一个人名是后一个人名链长加一，并不断更新最长链长。最后输出最长链长。

在网上站下来了一点map的用法

1. map最基本的构造函数；
   map<string , int >mapstring;         map<int ,string >mapint;
   map<sring, char>mapstring;         map< char ,string>mapchar;
   map<char ,int>mapchar;            map<int ,char >mapint；

2. map添加数据；

   map<int ,string> maplive;
   1.maplive.insert(pair<int,string>(102,"aclive"));
   2.maplive.insert(map<int,string>::value_type(321,"hai"));
   3, maplive[112]="April";//map中最简单最常用的插入添加！
3，map中元素的查找：

find()函数返回一个迭代器指向键值为key的元素，如果没找到就返回指向map尾部的迭代器。

   map<int ,string >::iterator l_it;;
   l_it=maplive.find(112);
   if(l_it==maplive.end())
                cout<<"we do not find 112"<<endl;
   else cout<<"wo find 112"<<endl;
4,map中元素的删除：
   如果删除112；
   map<int ,string >::iterator l_it;;
   l_it=maplive.find(112);
   if(l_it==maplive.end())
        cout<<"we do not find 112"<<endl;
   else  maplive.erase(l_it);  //delete 112;
5,map中 swap的用法：
  Map中的swap不是一个容器中的元素交换，而是两个容器交换；
  For example：
  #include <map>
  #include <iostream>

using namespace std;

  int main( )
  {
      map <int, int> m1, m2, m3;
      map <int, int>::iterator m1_Iter;

      m1.insert ( pair <int, int>  ( 1, 10 ) );
      m1.insert ( pair <int, int>  ( 2, 20 ) );
      m1.insert ( pair <int, int>  ( 3, 30 ) );
      m2.insert ( pair <int, int>  ( 10, 100 ) );
      m2.insert ( pair <int, int>  ( 20, 200 ) );
      m3.insert ( pair <int, int>  ( 30, 300 ) );

   cout << "The original map m1 is:";
   for ( m1_Iter = m1.begin( ); m1_Iter != m1.end( ); m1_Iter++ )
      cout << " " << m1_Iter->second;
      cout   << "." << endl;

   // This is the member function version of swap
   //m2 is said to be the argument map; m1 the target map
   m1.swap( m2 );

   cout << "After swapping with m2, map m1 is:";
   for ( m1_Iter = m1.begin( ); m1_Iter != m1.end( ); m1_Iter++ )
      cout << " " << m1_Iter -> second;
      cout  << "." << endl;
   cout << "After swapping with m2, map m2 is:";
   for ( m1_Iter = m2.begin( ); m1_Iter != m2.end( ); m1_Iter++ )
      cout << " " << m1_Iter -> second;
      cout  << "." << endl;
   // This is the specialized template version of swap
   swap( m1, m3 );

   cout << "After swapping with m3, map m1 is:";
   for ( m1_Iter = m1.begin( ); m1_Iter != m1.end( ); m1_Iter++ )
      cout << " " << m1_Iter -> second;
      cout   << "." << endl;
}

6.map的sort问题：
  Map中的元素是自动按key升序排序,所以不能对map用sort函数：
  For example：
  #include <map>
  #include <iostream>

using namespace std;

int main( )
{
map <int, int> m1;
map <int, int>::iterator m1_Iter;

   m1.insert ( pair <int, int>  ( 1, 20 ) );
   m1.insert ( pair <int, int>  ( 4, 40 ) );
   m1.insert ( pair <int, int>  ( 3, 60 ) );
   m1.insert ( pair <int, int>  ( 2, 50 ) );
   m1.insert ( pair <int, int>  ( 6, 40 ) );
   m1.insert ( pair <int, int>  ( 7, 30 ) );

   cout << "The original map m1 is:"<<endl;
   for ( m1_Iter = m1.begin( ); m1_Iter != m1.end( ); m1_Iter++ )
      cout <<  m1_Iter->first<<" "<<m1_Iter->second<<endl;

}
  The original map m1 is:
  1 20
  2 50
  3 60
  4 40
  6 40
  7 30
  请按任意键继续. . .

7，   map的基本操作函数：
      C++ Maps是一种关联式容器，包含“关键字/值”对
      begin()          返回指向map头部的迭代器
      clear(）         删除所有元素
      count()          返回指定元素出现的次数
      empty()          如果map为空则返回true
      end()            返回指向map末尾的迭代器
      equal_range()    返回特殊条目的迭代器对
      erase()          删除一个元素
      find()           查找一个元素
      get_allocator()  返回map的配置器
      insert()         插入元素
      key_comp()       返回比较元素key的函数
      lower_bound()    返回键值>=给定元素的第一个位置
      max_size()       返回可以容纳的最大元素个数
      rbegin()         返回一个指向map尾部的逆向迭代器
      rend()           返回一个指向map头部的逆向迭代器
      size()           返回map中元素的个数
      swap()            交换两个map
      upper_bound()     返回键值>给定元素的第一个位置
      value_comp()      返回比较元素value的函数

#include<iostream>
#include<cstdio>
#include<cstring>
#include<ctime>
#include<algorithm>
#include <map>
using namespace std;

//522A关于map

int main()
{
    map<string,int> s;//map是c++的一个标准容器，提供了一对一关系，
    s["polycarp"]=1;
    int n;
    cin>>n;
    int res=1;
    for(int i=0;i<n;i++)
    {
        string n1,n2;
        cin>>n1>>n2>>n2;
        for(int j=0;j<n1.length();j++)
        {
            n1[j]=tolower(n1[j]);
        }
        for(int j=0;j<n2.length();j++)
        {
            n2[j]=tolower(n2[j]);
        }
        s[n1]=s[n2]+1;
        res=max(res,s[n1]);
    }
    cout<<res<<endl;
    return 0;
}