哈密顿绕行世界问题
Time Limit : 3000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 93 Accepted Submission(s) : 52
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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
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string.h>
int a[21][3],st,c[21],ct;
bool b[21];
using namespace std;
int main()
{
void dfs(int, int);
memset(b, 0, sizeof(b));
for (int i = 1; i <= 20; i++)
{
int x, y, z;
scanf_s("%d%d%d", &x, &y, &z);
if (x > y) swap(x, y);
else if (x > z) swap(x, z);
if (y > z) swap(y, z);
a[i][0] = x, a[i][1] = y, a[i][2] = z;
}
while (scanf_s("%d", &st) != EOF)
{
if (st == 0) break;
ct = 0;
c[1] = st; b[st] = 1;
dfs(st, 2);
}
return 0;
}
void dfs(int de, int k)
{
for (int i = 0; i < 3; i++)
{
if (b[a[de][i]] == 0)
{
b[a[de][i]] = 1; c[k] = a[de][i];
if (k == 20)
{
if (a[c[k]][0] == st || a[c[k]][1] == st || a[c[k]][2] == st)
{
ct++;
printf("%d: ", ct);
for (int j= 1; j <= 20; j++)
printf("%d ", c[j]);
printf("%d\n", st);
}
b[c[k]] = 0;
return;
}
dfs(a[de][i], k + 1);
b[a[de][i]] = 0;
}
}
}
