10054 - The Necklace//欧拉回路

/*判断有无欧拉回路
  还是老样子：
  先判断图是否连通，这道题是无向图，然后判断每个点的度，最后打印之
*/
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 1005;
int G[maxn][maxn];
int degree[maxn];
int f[maxn];
int n;
int find(int x)
{
    if(x != f[x])
    {
        f[x] = find(f[x]);
    }
    return f[x];
}
void euler(int u)
{
    for(int v = 1;v <= 50;v++)
    {
        if(G[u][v])
        {
            G[u][v]--;
            G[v][u]--;
            euler(v);
            printf("%d %d\n",v,u);
        }
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    for(int t = 1;t <= T;t++)
    {
        memset(G,0,sizeof(G));
        memset(degree,0,sizeof(degree));
        for(int i = 1;i <= 50;i++) f[i] = i;

        scanf("%d",&n);
        for(int i = 1;i <= n;i++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            G[a][b]++;
            G[b][a]++;
            degree[a]++;
            degree[b]++;
            if( find(a) != find(b))
            {
                f[find(a)] = find(b);
            }
        }
        int j;
        bool ok = true;
        for(j = 0;!degree[j];j++);
        for(int i = j + 1;i < maxn;i++)
        {
            if(degree[i] && find(j) != find(i))
            {
                ok = false;
                break;
            }
        }
        printf("Case #%d\n",t);
        if(ok)
        {
            for(int i = 1;i < maxn;i++)
            {
                if(degree[i]%2)
                {
                    ok = false;
                    break;
                }
            }
            if(ok)
            {
                for(j = 0;!degree[j];j++);
                euler(j);
            }
        }
        if(!ok) printf("some beads may be lost\n");
        printf("\n");
    }
    return 0;
}