HDU2181(KB2-C)
哈密顿绕行世界问题
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3890 Accepted Submission(s): 2358
Problem Description
一个规则的实心十二面体,它的 20个顶点标出世界著名的20个城市,你从一个城市出发经过每个城市刚好一次后回到出发的城市。
Input
前20行的第i行有3个数,表示与第i个城市相邻的3个城市.第20行以后每行有1个数m,m<=20,m>=1.m=0退出.
Output
输出从第m个城市出发经过每个城市1次又回到m的所有路线,如有多条路线,按字典序输出,每行1条路线.每行首先输出是第几条路线.然后个一个: 后列出经过的城市.参看Sample output
Sample Input
2 5 20
1 3 12
2 4 10
3 5 8
1 4 6
5 7 19
6 8 17
4 7 9
8 10 16
3 9 11
10 12 15
2 11 13
12 14 20
13 15 18
11 14 16
9 15 17
7 16 18
14 17 19
6 18 20
1 13 19
5
0
Sample Output
1: 5 1 2 3 4 8 7 17 18 14 15 16 9 10 11 12 13 20 19 6 5
2: 5 1 2 3 4 8 9 10 11 12 13 20 19 18 14 15 16 17 7 6 5
3: 5 1 2 3 10 9 16 17 18 14 15 11 12 13 20 19 6 7 8 4 5
4: 5 1 2 3 10 11 12 13 20 19 6 7 17 18 14 15 16 9 8 4 5
5: 5 1 2 12 11 10 3 4 8 9 16 15 14 13 20 19 18 17 7 6 5
6: 5 1 2 12 11 15 14 13 20 19 18 17 16 9 10 3 4 8 7 6 5
7: 5 1 2 12 11 15 16 9 10 3 4 8 7 17 18 14 13 20 19 6 5
8: 5 1 2 12 11 15 16 17 18 14 13 20 19 6 7 8 9 10 3 4 5
9: 5 1 2 12 13 20 19 6 7 8 9 16 17 18 14 15 11 10 3 4 5
10: 5 1 2 12 13 20 19 18 14 15 11 10 3 4 8 9 16 17 7 6 5
11: 5 1 20 13 12 2 3 4 8 7 17 16 9 10 11 15 14 18 19 6 5
12: 5 1 20 13 12 2 3 10 11 15 14 18 19 6 7 17 16 9 8 4 5
13: 5 1 20 13 14 15 11 12 2 3 10 9 16 17 18 19 6 7 8 4 5
14: 5 1 20 13 14 15 16 9 10 11 12 2 3 4 8 7 17 18 19 6 5
15: 5 1 20 13 14 15 16 17 18 19 6 7 8 9 10 11 12 2 3 4 5
16: 5 1 20 13 14 18 19 6 7 17 16 15 11 12 2 3 10 9 8 4 5
17: 5 1 20 19 6 7 8 9 10 11 15 16 17 18 14 13 12 2 3 4 5
18: 5 1 20 19 6 7 17 18 14 13 12 2 3 10 11 15 16 9 8 4 5
19: 5 1 20 19 18 14 13 12 2 3 4 8 9 10 11 15 16 17 7 6 5
20: 5 1 20 19 18 17 16 9 10 11 15 14 13 12 2 3 4 8 7 6 5
21: 5 4 3 2 1 20 13 12 11 10 9 8 7 17 16 15 14 18 19 6 5
22: 5 4 3 2 1 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5
23: 5 4 3 2 12 11 10 9 8 7 6 19 18 17 16 15 14 13 20 1 5
24: 5 4 3 2 12 13 14 18 17 16 15 11 10 9 8 7 6 19 20 1 5
25: 5 4 3 10 9 8 7 6 19 20 13 14 18 17 16 15 11 12 2 1 5
26: 5 4 3 10 9 8 7 17 16 15 11 12 2 1 20 13 14 18 19 6 5
27: 5 4 3 10 11 12 2 1 20 13 14 15 16 9 8 7 17 18 19 6 5
28: 5 4 3 10 11 15 14 13 12 2 1 20 19 18 17 16 9 8 7 6 5
29: 5 4 3 10 11 15 14 18 17 16 9 8 7 6 19 20 13 12 2 1 5
30: 5 4 3 10 11 15 16 9 8 7 17 18 14 13 12 2 1 20 19 6 5
31: 5 4 8 7 6 19 18 17 16 9 10 3 2 12 11 15 14 13 20 1 5
32: 5 4 8 7 6 19 20 13 12 11 15 14 18 17 16 9 10 3 2 1 5
33: 5 4 8 7 17 16 9 10 3 2 1 20 13 12 11 15 14 18 19 6 5
34: 5 4 8 7 17 18 14 13 12 11 15 16 9 10 3 2 1 20 19 6 5
35: 5 4 8 9 10 3 2 1 20 19 18 14 13 12 11 15 16 17 7 6 5
36: 5 4 8 9 10 3 2 12 11 15 16 17 7 6 19 18 14 13 20 1 5
37: 5 4 8 9 16 15 11 10 3 2 12 13 14 18 17 7 6 19 20 1 5
38: 5 4 8 9 16 15 14 13 12 11 10 3 2 1 20 19 18 17 7 6 5
39: 5 4 8 9 16 15 14 18 17 7 6 19 20 13 12 11 10 3 2 1 5
40: 5 4 8 9 16 17 7 6 19 18 14 15 11 10 3 2 12 13 20 1 5
41: 5 6 7 8 4 3 2 12 13 14 15 11 10 9 16 17 18 19 20 1 5
42: 5 6 7 8 4 3 10 9 16 17 18 19 20 13 14 15 11 12 2 1 5
43: 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 2 3 4 5
44: 5 6 7 8 9 16 17 18 19 20 1 2 12 13 14 15 11 10 3 4 5
45: 5 6 7 17 16 9 8 4 3 10 11 15 14 18 19 20 13 12 2 1 5
46: 5 6 7 17 16 15 11 10 9 8 4 3 2 12 13 14 18 19 20 1 5
47: 5 6 7 17 16 15 11 12 13 14 18 19 20 1 2 3 10 9 8 4 5
48: 5 6 7 17 16 15 14 18 19 20 13 12 11 10 9 8 4 3 2 1 5
49: 5 6 7 17 18 19 20 1 2 3 10 11 12 13 14 15 16 9 8 4 5
50: 5 6 7 17 18 19 20 13 14 15 16 9 8 4 3 10 11 12 2 1 5
51: 5 6 19 18 14 13 20 1 2 12 11 15 16 17 7 8 9 10 3 4 5
52: 5 6 19 18 14 15 11 10 9 16 17 7 8 4 3 2 12 13 20 1 5
53: 5 6 19 18 14 15 11 12 13 20 1 2 3 10 9 16 17 7 8 4 5
54: 5 6 19 18 14 15 16 17 7 8 9 10 11 12 13 20 1 2 3 4 5
55: 5 6 19 18 17 7 8 4 3 2 12 11 10 9 16 15 14 13 20 1 5
56: 5 6 19 18 17 7 8 9 16 15 14 13 20 1 2 12 11 10 3 4 5
57: 5 6 19 20 1 2 3 10 9 16 15 11 12 13 14 18 17 7 8 4 5
58: 5 6 19 20 1 2 12 13 14 18 17 7 8 9 16 15 11 10 3 4 5
59: 5 6 19 20 13 12 11 10 9 16 15 14 18 17 7 8 4 3 2 1 5
60: 5 6 19 20 13 14 18 17 7 8 4 3 10 9 16 15 11 12 2 1 5
Author
Zhousc
Source
裸的dfs。
1 //2017-03-01 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 #include <algorithm> 6 7 using namespace std; 8 9 int M[20][5], ans[25], vis[25], m, kase; 10 11 void dfs(int city, int step) 12 { 13 if(step == 20 && city == m){ 14 cout<<++kase<<": "; 15 for(int i = 0; i < 20; i++) 16 cout<<ans[i]<<" "; 17 cout<<m<<endl; 18 return; 19 } 20 ans[step] = city; 21 for(int i = 0; i < 3; i++) 22 { 23 if(!vis[M[city][i]]) 24 { 25 vis[M[city][i]] = 1; 26 dfs(M[city][i], step+1); 27 vis[M[city][i]] = 0; 28 } 29 } 30 } 31 32 int main() 33 { 34 while(cin>>M[1][0]>>M[1][1]>>M[1][2]) 35 { 36 sort(M[0], M[0]+3); 37 for(int i = 2; i <= 20; i++) 38 { 39 cin>>M[i][0]>>M[i][1]>>M[i][2]; 40 sort(M[i], M[i]+3); 41 } 42 while(cin>>m && m) 43 { 44 memset(vis, 0, sizeof(vis)); 45 kase = 0; 46 dfs(m, 0); 47 } 48 } 49 50 return 0; 51 }