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

2

3 3

1 2 2 2

2 3 2 2

3 1 2 2

3 3

1 2 10 2

2 3 2 2

3 1 2 2

 

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.

题意

给你N个点M条边强连通的有向简单图，D代表删掉这个边的花费，D+B代表重建为双向边的花费，让你选择一个集合S，其余的点在T集合，X为u在S集合v在T集合的所有边的D之和，Y为u在T集合v在S集合的所有边的D+B之和，求是否存在一个集合S，使得X>Y，若存在输出unhappy，否则输出happy

题解

无源汇上下界网络流，下界D，上界D+B，判断是否存在可行流

若存在，则说明对于任意集合S，流出的流量=流入的流量，X<=流出的流量<=Y

建图每条边建为自由流（u，v，B）

对于每个点，设M为总流入-总流出

若M>0，则建（S，i，M）说明i需要多流出M

若M<0，则建（i，T，M）说明i需要多流入M

最后判断与S连的边是否全满流

代码

 

1 #include
     
       2 using namespace std; 3  4 const int maxn=1e5+5; 5 const int maxm=2e5+5; 6 const int INF=0x3f3f3f3f; 7  8 int TO[maxm],CAP[maxm],NEXT[maxm],tote; 9 int FIR[maxn],gap[maxn],cur[maxn],d[maxn],q[400000];10 int n,m,S,T;11 12 void add(int u,int v,int cap)13 {14     TO[tote]=v;15     CAP[tote]=cap;16     NEXT[tote]=FIR[u];17     FIR[u]=tote++;18     19     TO[tote]=u;20     CAP[tote]=0;21     NEXT[tote]=FIR[v];22     FIR[v]=tote++;23 }24 void bfs()25 {26     memset(gap,0,sizeof gap);27     memset(d,0,sizeof d);28     ++gap[d[T]=1];29     for(int i=1;i<=n;++i)cur[i]=FIR[i];30     int head=1,tail=1;31     q[1]=T;32     while(head<=tail)33     {34         int u=q[head++];35         for(int v=FIR[u];v!=-1;v=NEXT[v])36             if(!d[TO[v]])37                 ++gap[d[TO[v]]=d[u]+1],q[++tail]=TO[v];38     }39 }40 int dfs(int u,int fl)41 {42     if(u==T)return fl;43     int flow=0;44     for(int &v=cur[u];v!=-1;v=NEXT[v])45         if(CAP[v]&&d[u]==d[TO[v]]+1)46         {47             int Min=dfs(TO[v],min(fl,CAP[v]));48             flow+=Min,fl-=Min,CAP[v]-=Min,CAP[v^1]+=Min;49             if(!fl)return flow;50         }51     if(!(--gap[d[u]]))d[S]=n+1;52     ++gap[++d[u]],cur[u]=FIR[u];53     return flow;54 }55 int ISAP()56 {57     bfs();58     int ret=0;59     while(d[S]<=n)ret+=dfs(S,INF);60     return ret;61 }62 void init()63 {64     tote=0;65     memset(FIR,-1,sizeof FIR);66 }67 int in[maxn];68 int main()69 {70     int t;71     scanf("%d",&t);72     for(int ca=1;ca<=t;ca++)73     {74         init();75         memset(in,0,sizeof in);76         scanf("%d%d",&n,&m);77         for(int i=0,u,v,d,b;i
      
       0)88             {89                 add(S,i,in[i]);90                 sum+=in[i];91             }92             else if(in[i]<0)93                 add(i,T,-in[i]);94         printf("Case #%d: %s\n",ca,ISAP()==sum?"happy":"unhappy");95     }96     return 0;97 }