Reposts（dfs）

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.

Examples
Input
5
tourist reposted Polycarp
Petr reposted Tourist
WJMZBMR reposted Petr
sdya reposted wjmzbmr
vepifanov reposted sdya
Output
6
題目分析：
這道題只要一開始全部字母處理成小寫或大寫，數字和‘/’不用管的。還有一點要注意的是他們轉發的順序可能不是連續的，可能隔了好多個人又轉發了一連串，只要注意這一點就行了。深搜很輕鬆。
代碼：

#include<iostream>
#include<string>
using namespace std;
int shu[201],n;
string a[201],b,c[201];
int dfs(string ok,int len,int st)
{
    int l=len;
    for(int i=st+1;i<=n;i++)
    {
        if(c[i]==ok)
        {
            l=max(l,dfs(a[i],len+1,i));//取最大值
        }
    }
    return l;
}
int main()
{
    int ans=1;
    cin>>n;
    string t="polycarp";
    int l=1;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        for(int j=0;j<a[i].size();j++)
            if(a[i][j]>='A'&&a[i][j]<='Z') a[i][j]+=32;
        cin>>b;
        cin>>c[i];
        for(int j=0;j<c[i].size();j++)
            if(c[i][j]>='A'&&c[i][j]<='Z') c[i][j]+=32;
        if(c[i]==t) {
            shu[l++]=i;
        }
    }
    int len;
    string ok;
    for(int i=1;i<=l-1;i++)
    {
        len=2;
        ok=a[shu[i]];
        ans=max(dfs(ok,len,shu[i]),ans);//取最大值
    }
    cout<<ans;
    return 0;
}