哈密顿绕行世界问题 HDU - 2181 DFS

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): 4900 Accepted Submission(s): 2951

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

ECJTU 2008 Summer Contest

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <queue>
#include <algorithm>
using namespace std;
int w[25][3];
int book[24];
int temp[24];
int m;
//int dir[4][2]={{0,1}, {1,0}, {0,-1}, {-1,0}};
int o=0;
void dfs(int x, int s)
{
if(s==20)
{
printf("%d: ",++o,m);
for(int i=0;i<=s;i++)
{
printf(" %d",temp[i]);
}
printf("\n");
return ;
}
for(int i=0;i<3;i++)
{
int tx = w[x][i];
if(!book[tx])
{
book[tx] = 1;
temp[s+1] = tx;//因为temp[0]已经存了m,所以存s+1里
dfs(tx, s+1);
book[tx] = 0;
}
if(s==19 && tx==m)
{
temp[s+1] = m;//最后一个也存m
dfs(m,s+1);
}
}
}

int main()
{
for(int i=1;i<=20;i++)
{
scanf("%d%d%d",&w[i][0],&w[i][1],&w[i][2]);
sort(w[i],w[i]+3);
}
while(cin>>m,m!=0)
{
memset(book,0,sizeof(book));
book[m]=1;
o=0;
temp[0]=m;//第一个存m
dfs(m, 0);
}
return 0;
}