VK Cup 2015 - Qualification Round 1 A. Reposts [ dp DAG上最长路 ]

传送门

A. Reposts

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

6

Input

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

Output

2

Input

1 SoMeStRaNgEgUe reposted PoLyCaRp

Output

2

题解：
DAG上最长路，用记忆化dp解决

核心代码

 1 int dfs(int now)
 2 {
 3     int i;
 4     if(dp[now]>0) return dp[now];
 5     dp[now]=1;
 6     for(i=0;i<G[now].size();i++){
 7         dp[now]=max(dp[now],dfs(G[now][i])+1);
 8     }
 9     return dp[now];
10 }

 

10702991

2015-04-14 14:15:05

njczy2010

A - Reposts

GNU C++

Accepted

15 ms

40 KB

 

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <stack>
 4 #include <vector>
 5 #include <algorithm>
 6 #include <queue>
 7 #include <map>
 8 #include <string>
 9 
10 #define ll long long
11 int const N = 405;
12 int const M = 205;
13 int const INF = 0x3f3f3f3f;
14 ll const mod = 1000000007;
15 
16 using namespace std;
17 
18 int n;
19 char s1[N],s2[N],te[N];
20 map<string,int> mp;
21 int tot;
22 int ans;
23 vector<int> G[N];
24 int dp[N];
25 
26 void ini()
27 {
28     memset(dp,0,sizeof(dp));
29     ans=tot=0;
30     int i,j,l1,l2;
31     mp.clear();
32     for(i=0;i<=200;i++){
33         G[i].clear();
34     }
35     for(i=1;i<=n;i++){
36         scanf("%s%s%s",s2,te,s1);
37         l1=strlen(s1);
38         l2=strlen(s2);
39         for(j=0;j<l1;j++){
40             if(s1[j]>='A' && s1[j]<='Z'){
41                 s1[j]=s1[j]-'A'+'a';
42             }
43         }
44         for(j=0;j<l2;j++){
45             if(s2[j]>='A' && s2[j]<='Z'){
46                 s2[j]=s2[j]-'A'+'a';
47             }
48         }
49         if(mp.count(s1)==0){
50             tot++;
51             mp[s1]=tot;
52         }
53         if(mp.count(s2)==0){
54             tot++;
55             mp[s2]=tot;
56         }
57         G[ mp[s1] ].push_back( mp[s2] );
58     }
59 }
60 
61 int dfs(int now)
62 {
63     int i;
64     if(dp[now]>0) return dp[now];
65     dp[now]=1;
66     for(i=0;i<G[now].size();i++){
67         dp[now]=max(dp[now],dfs(G[now][i])+1);
68     }
69     return dp[now];
70 }
71 
72 void solve()
73 {
74     ans=dfs(1);
75 }
76 
77 void out()
78 {
79     printf("%d\n",ans);
80 }
81 
82 int main()
83 {
84     //freopen("data.in","r",stdin);
85     //scanf("%d",&T);
86     //for(int cnt=1;cnt<=T;cnt++)
87     //while(T--)
88     while(scanf("%d",&n)!=EOF)
89     {
90         ini();
91         solve();
92         out();
93     }
94 }