将每个颜色看成一个顶点，对于每个珠子在两个颜色之间连一条无向边，然后求欧拉回路。
1 #include <cstdio> 2 #include <cstring> 3 4 const int maxn = 50 + 5; 5 int G[maxn][maxn], deg[maxn]; 6 7 void Euler(int u) 8 { 9 for(int v = 1; v <= 50; v++) if(G[u][v]) 10 { 11 G[u][v]--; G[v][u]--; 12 Euler(v); 13 printf("%d %d ", v, u); 14 } 15 } 16 17 int main() 18 { 19 //freopen("in.txt", "r", stdin); 20 21 int T; scanf("%d", &T); 22 for(int kase = 1; kase <= T; kase++) 23 { 24 memset(G, 0, sizeof(G)); 25 memset(deg, 0, sizeof(deg)); 26 27 if(kase > 1) puts(""); 28 printf("Case #%d ", kase); 29 30 int n, a, b, st; 31 scanf("%d", &n); 32 for(int i = 0; i < n; i++) 33 { 34 scanf("%d%d", &a, &b); 35 st = a; 36 deg[a]++; deg[b]++; 37 G[a][b]++; G[b][a]++; 38 } 39 40 bool ok = true; 41 for(int i = 1; i <= 50; i++) if(deg[i] % 2) { ok = false; break; } 42 if(!ok) { puts("some beads may be lost"); continue; } 43 44 Euler(st); 45 46 } 47 48 return 0; 49 }