tjut 4975

#include <queue>
#include <vector>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <iostream>
using namespace std;
const int INF=0x3f3f3f3f;
struct node{
int u,v,cap;
node(){}
node(int u,int v,int cap):u(u),v(v),cap(cap){}
}es[1000005];
int R,S,T;
int dis[50005],vis[50005],iter[50005];
vector<int> tab[50005];
vector<int> G[50005];
void addedge(int u, int v, int cap){
tab[u].push_back(R);
es[R++]=node(u,v,cap);
tab[v].push_back(R);
es[R++]=node(v,u,0);
}
int bfs(){
int i,h;
queue<int> q;
q.push(S);
memset(dis,INF,sizeof(dis));
dis[S]=0;
while(q.size()){
h=q.front();
q.pop();
for(i=0;i<tab[h].size();i++){
node &e=es[tab[h][i]];
if(e.cap>0&&dis[e.v]==INF){
dis[e.v]=dis[h]+1;
q.push(e.v);
}
}
}
return dis[T]<INF;
}
int dfs(int x,int maxflow){
int flow;
if(x==T)
return maxflow;
for(int &i=iter[x];i<tab[x].size();i++){
node &e=es[tab[x][i]];
if(dis[e.v]==dis[x]+1&&e.cap>0){
flow=dfs(e.v,min(maxflow,e.cap));
if(flow){
e.cap-=flow;
es[tab[x][i]^1].cap+=flow;
return flow;
}
}
}
return 0;
}
int dinic(){
int ans,flow;
ans=0;
while(bfs()){
memset(iter,0,sizeof(iter));
while(flow=dfs(S,INF))
ans+=flow;
}
return ans;
}                                               //dinic模板
int judge(int s,int t){
int i,tmp;
vis[s]=1;
for(i=0;i<G[s].size();i++){
tmp=G[s][i];
if(tmp==t)                              //防止出现两个节点的环
continue;
if(vis[tmp]==1)
return 1;
if(judge(tmp,s))
return 1;
}
vis[s]=0;
return 0;
}
int main(){
int N,M,i,j,l,r,t,ans,cas,num,sign;
cas=1;
scanf("%d",&t);
while(t--){                                 //以行和列建立二分图跑网络流
scanf("%d%d",&N,&M);
l=r=0;
S=0,R=0,T=N+M+1;
for(i=0;i<=T;i++){
G[i].clear();
tab[i].clear();
}
memset(vis,0,sizeof(vis));
for(i=1;i<=N;i++){
scanf("%d",&num);
addedge(S,i,num);
l+=num;
}
for(i=1;i<=M;i++){
scanf("%d",&num);
addedge(i+N,T,num);
r+=num;
}
for(i=1;i<=N;i++)
for(j=1;j<=M;j++)
addedge(i,j+N,9);                       //因为只能0~9所以容量建为9
if(l!=r){                               //如果行的和和列的和不相等则一定无解
printf("Case #%d: So naive!\n",cas++);
continue;
}
ans=dinic();
if(ans!=l){                             //不是满流的也无解
printf("Case #%d: So naive!\n",cas++);
continue;
}
for(i=0;i<R;i++)
if(es[i].cap>0)                         //将残余网络保存下来判断是否有环，从而
G[es[i].u].push_back(es[i].v);          //确定是否有多组解，因为当残余网络中有
sign=0;                                 //环时则可以通过残余网络去跑一些原来的流量
for(i=1;i<=N;i++)
if(judge(i,-1)){
sign=1;
break;
}
if(sign)
printf("Case #%d: So young!\n",cas++);
else
printf("Case #%d: So simple!\n",cas++);
}
return 0;
}