Problem Description

Tom is a commander, his task is destroying his enemy’s transportation system.

Let’s represent his enemy’s transportation system as a simple
directed graph G with n nodes and m edges. Each node is a city and each
directed edge is a directed road. Each edge from node u to node v is
associated with two values D and B, D is the cost to destroy/remove such
edge, B is the cost to build an undirected edge between u and v.

His enemy can deliver supplies from city u to city v if and only if
there is a directed path from u to v. At first they can deliver supplies
from any city to any other cities. So the graph is a strongly-connected
graph.

He will choose a non-empty proper subset of cities,
let’s denote this set as S. Let’s denote the complement set of S as T.
He will command his soldiers to destroy all the edges (u, v) that u
belongs to set S and v belongs to set T.

To destroy an edge, he must pay the related cost D. The total cost he will pay is X. You can use this formula to calculate X:

After that, all the edges from S to T are destroyed. In order to
deliver huge number of supplies from S to T, his enemy will change all
the remained directed edges (u, v) that u belongs to set T and v belongs
to set S into undirected edges. (Surely, those edges exist because the
original graph is strongly-connected)

To change an edge, they
must remove the original directed edge at first, whose cost is D, then
they have to build a new undirected edge, whose cost is B. The total
cost they will pay is Y. You can use this formula to calculate Y:

At last, if Y>=X, Tom will achieve his goal. But Tom is so lazy
that he is unwilling to take a cup of time to choose a set S to make
Y>=X, he hope to choose set S randomly! So he asks you if there is a
set S, such that Y<X. If such set exists, he will feel unhappy,
because he must choose set S carefully, otherwise he will become very
happy.

 

Input

There are multiply test cases.

The first line contains an integer T(T<=200), indicates the number of cases.

For each test case, the first line has two numbers n and m.

Next m lines describe each edge. Each line has four numbers u, v, D, B.
(2=<n<=200, 2=<m<=5000, 1=<u, v<=n, 0=<D, B<=100000)

The meaning of all characters are described above. It is guaranteed that the input graph is strongly-connected.

 

Output

For each case, output "Case #X: " first, X is the case number
starting from 1.If such set doesn’t exist, print “happy”, else print
“unhappy”.

 

Sample Input

 

Sample Output

Case #1: happy

Case #2: unhappy

 

Sample Output

In first sample, for any set S, X=2, Y=4.

In second sample. S= {1}, T= {2, 3}, X=10, Y=4.

 #include<bits/stdc++.h>

 using namespace std;

 const int maxn=1e5+;

 const int maxm=2e5+;

 const int INF=0x3f3f3f3f;

 int TO[maxm],CAP[maxm],NEXT[maxm],tote;

 int FIR[maxn],gap[maxn],cur[maxn],d[maxn],q[];

 int n,m,S,T;

 void add(int u,int v,int cap)

 {

     TO[tote]=v;

     CAP[tote]=cap;

     NEXT[tote]=FIR[u];

     FIR[u]=tote++;

     TO[tote]=u;

     CAP[tote]=;

     NEXT[tote]=FIR[v];

     FIR[v]=tote++;

 }

 void bfs()

 {

     memset(gap,,sizeof gap);

     memset(d,,sizeof d);

     ++gap[d[T]=];

     for(int i=;i<=n;++i)cur[i]=FIR[i];

     int head=,tail=;

     q[]=T;

     while(head<=tail)

     {

         int u=q[head++];

         for(int v=FIR[u];v!=-;v=NEXT[v])

             if(!d[TO[v]])

                 ++gap[d[TO[v]]=d[u]+],q[++tail]=TO[v];

     }

 }

 int dfs(int u,int fl)

 {

     if(u==T)return fl;

     int flow=;

     for(int &v=cur[u];v!=-;v=NEXT[v])

         if(CAP[v]&&d[u]==d[TO[v]]+)

         {

             int Min=dfs(TO[v],min(fl,CAP[v]));

             flow+=Min,fl-=Min,CAP[v]-=Min,CAP[v^]+=Min;

             if(!fl)return flow;

         }

     if(!(--gap[d[u]]))d[S]=n+;

     ++gap[++d[u]],cur[u]=FIR[u];

     return flow;

 }

 int ISAP()

 {

     bfs();

     int ret=;

     while(d[S]<=n)ret+=dfs(S,INF);

     return ret;

 }

 void init()

 {

     tote=;

     memset(FIR,-,sizeof FIR);

 }

 int in[maxn];

 int main()

 {

     int t;

     scanf("%d",&t);

     for(int ca=;ca<=t;ca++)

     {

         init();

         memset(in,,sizeof in);

         scanf("%d%d",&n,&m);

         for(int i=,u,v,d,b;i<m;i++)

         {

             scanf("%d%d%d%d",&u,&v,&d,&b);

             add(u,v,b);

             in[u]-=d;

             in[v]+=d;

         }

         S=n+,T=S+,n+=;

         int sum=;

         for(int i=;i<=n;i++)

             if(in[i]>)

             {

                 add(S,i,in[i]);

                 sum+=in[i];

             }

             else if(in[i]<)

                 add(i,T,-in[i]);

         printf("Case #%d: %s\n",ca,ISAP()==sum?"happy":"unhappy");

     }

     return ;

 }