Loj 1003–Drunk(拓扑排序)

1003 - Drunk

Time Limit: 2 second(s)
Memory Limit: 32 MB

One of my friends is always drunk. So, sometimes I get a bit confused whether he is drunk or not. So, one day I was talking to him, about his drinks! He began to describe his way of drinking. So, let me share his ideas a bit. I am expressing in my words.

There are many kinds of drinks, which he used to take. But there are some rules; there are some drinks that have some pre requisites. Suppose if you want to take wine, you should have taken soda, water before it. That's why to get real drunk is not that easy.

Now given the name of some drinks! And the prerequisites of the drinks, you have to say that whether it's possible to get drunk or not. To get drunk, a person should take all the drinks.

Input

Input starts with an integer T (≤ 50), denoting the number of test cases.

Each case starts with an integer m (1 ≤ m ≤ 10000). Each of the next m lines will contain two names each in the format a b, denoting that you must have a before having b. The names will contain at most10 characters with no blanks.

Output

For each case, print the case number and 'Yes' or 'No', depending on whether it's possible to get drunk or not.

Sample Input

soda wine

water wine

soda wine

water wine

wine water

Output for Sample Input

Case 1: Yes

Case 2: No

：：学习一个新算法，总没那么容易，今天又因为一个弄错一个字母找了半天*.*!!

   1: #include <iostream>

   2: #include <cstdio>

   3: #include <algorithm>

   4: #include <cstring>

   5: #include <map>

   6: using namespace std;

   7: const int maxn=11000;

   8: int head[maxn],in[maxn];

   9: bool vis[maxn];

  10: int ant,cas=1,id;

11:

  12: map<string,int>a;

  13: struct EDGE{

  14:     int v, next;

  15:     EDGE(){}

  16:     EDGE(int _v, int _next){v = _v, next = _next;}

  17: }e[maxn];

  18: int ecnt;

19:

  20: void add(int u, int v){

  21:     e[ecnt] = EDGE(v, head[u]);

  22:     head[u] = ecnt++;

  23: }

24:

  25: bool topo(){

  26:     memset(vis, 0, sizeof(vis));

27:

  28:     for(int t = 0; t < id; t++){

  29:         int u;

  30:         for(u = 0; u < id; u++)

  31:             if(!vis[u] && !in[u])

  32:                 break;

  33:         if(u >= id)

  34:             return false;

35:

  36:         vis[u] = true;

  37:         in[u]--;

  38:         for(int i = head[u]; i != -1; i = e[i].next){

  39:             int v = e[i].v;

  40:             in[v]--;

  41:         }

  42:     }

  43:     return true;

  44: }

45:

  46: void solve()

  47: {

  48:     int n;

  49:     ecnt=0,id=0;

  50:     a.clear();

  51:     memset(head,-1,sizeof(head));

  52:     memset(in,0,sizeof(in));

  53:     cin>>n;

54:

  55:     while(n--)

  56:     {

  57:         string x,y;

  58:         cin>>x>>y;

59:

  60:         if(a.find(x)==a.end()) a[x]=id++;

  61:         if(a.find(y)==a.end()) a[y]=id++;

  62:         int u=a[x],v=a[y];

  63:         add(u,v);

  64:         in[v]++;

  65:     }

  66:     if(topo())

  67:         cout<<"Yes"<<endl;

  68:     else

  69:         cout<<"No"<<endl;

  70: }

71:

  72: int main()

  73: {

  74:     ios::sync_with_stdio(false);

  75:     int t, cas = 0;

  76:     cin>>t;

  77:     while(t--){

  78:         cout<<"Case "<< (++cas)<<": ";

  79:         solve();

  80:     }

  81:     return 0;

  82: }