HDU-2181-哈密顿绕行世界问题

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题目信息：

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): 2628 Accepted Submission(s): 1640

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 5 201 3 122 4 103 5 81 4 65 7 196 8 174 7 98 10 163 9 1110 12 152 11 1312 14 2013 15 1811 14 169 15 177 16 1814 17 196 18 201 13 1950

Sample Output

1:  5 1 2 3 4 8 7 17 18 14 15 16 9 10 11 12 13 20 19 6 52:  5 1 2 3 4 8 9 10 11 12 13 20 19 18 14 15 16 17 7 6 53:  5 1 2 3 10 9 16 17 18 14 15 11 12 13 20 19 6 7 8 4 54:  5 1 2 3 10 11 12 13 20 19 6 7 17 18 14 15 16 9 8 4 55:  5 1 2 12 11 10 3 4 8 9 16 15 14 13 20 19 18 17 7 6 56:  5 1 2 12 11 15 14 13 20 19 18 17 16 9 10 3 4 8 7 6 57:  5 1 2 12 11 15 16 9 10 3 4 8 7 17 18 14 13 20 19 6 58:  5 1 2 12 11 15 16 17 18 14 13 20 19 6 7 8 9 10 3 4 59:  5 1 2 12 13 20 19 6 7 8 9 16 17 18 14 15 11 10 3 4 510:  5 1 2 12 13 20 19 18 14 15 11 10 3 4 8 9 16 17 7 6 511:  5 1 20 13 12 2 3 4 8 7 17 16 9 10 11 15 14 18 19 6 512:  5 1 20 13 12 2 3 10 11 15 14 18 19 6 7 17 16 9 8 4 513:  5 1 20 13 14 15 11 12 2 3 10 9 16 17 18 19 6 7 8 4 514:  5 1 20 13 14 15 16 9 10 11 12 2 3 4 8 7 17 18 19 6 515:  5 1 20 13 14 15 16 17 18 19 6 7 8 9 10 11 12 2 3 4 516:  5 1 20 13 14 18 19 6 7 17 16 15 11 12 2 3 10 9 8 4 517:  5 1 20 19 6 7 8 9 10 11 15 16 17 18 14 13 12 2 3 4 518:  5 1 20 19 6 7 17 18 14 13 12 2 3 10 11 15 16 9 8 4 519:  5 1 20 19 18 14 13 12 2 3 4 8 9 10 11 15 16 17 7 6 520:  5 1 20 19 18 17 16 9 10 11 15 14 13 12 2 3 4 8 7 6 521:  5 4 3 2 1 20 13 12 11 10 9 8 7 17 16 15 14 18 19 6 522:  5 4 3 2 1 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 523:  5 4 3 2 12 11 10 9 8 7 6 19 18 17 16 15 14 13 20 1 524:  5 4 3 2 12 13 14 18 17 16 15 11 10 9 8 7 6 19 20 1 525:  5 4 3 10 9 8 7 6 19 20 13 14 18 17 16 15 11 12 2 1 526:  5 4 3 10 9 8 7 17 16 15 11 12 2 1 20 13 14 18 19 6 527:  5 4 3 10 11 12 2 1 20 13 14 15 16 9 8 7 17 18 19 6 528:  5 4 3 10 11 15 14 13 12 2 1 20 19 18 17 16 9 8 7 6 529:  5 4 3 10 11 15 14 18 17 16 9 8 7 6 19 20 13 12 2 1 530:  5 4 3 10 11 15 16 9 8 7 17 18 14 13 12 2 1 20 19 6 531:  5 4 8 7 6 19 18 17 16 9 10 3 2 12 11 15 14 13 20 1 532:  5 4 8 7 6 19 20 13 12 11 15 14 18 17 16 9 10 3 2 1 533:  5 4 8 7 17 16 9 10 3 2 1 20 13 12 11 15 14 18 19 6 534:  5 4 8 7 17 18 14 13 12 11 15 16 9 10 3 2 1 20 19 6 535:  5 4 8 9 10 3 2 1 20 19 18 14 13 12 11 15 16 17 7 6 536:  5 4 8 9 10 3 2 12 11 15 16 17 7 6 19 18 14 13 20 1 537:  5 4 8 9 16 15 11 10 3 2 12 13 14 18 17 7 6 19 20 1 538:  5 4 8 9 16 15 14 13 12 11 10 3 2 1 20 19 18 17 7 6 539:  5 4 8 9 16 15 14 18 17 7 6 19 20 13 12 11 10 3 2 1 540:  5 4 8 9 16 17 7 6 19 18 14 15 11 10 3 2 12 13 20 1 541:  5 6 7 8 4 3 2 12 13 14 15 11 10 9 16 17 18 19 20 1 542:  5 6 7 8 4 3 10 9 16 17 18 19 20 13 14 15 11 12 2 1 543:  5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 2 3 4 544:  5 6 7 8 9 16 17 18 19 20 1 2 12 13 14 15 11 10 3 4 545:  5 6 7 17 16 9 8 4 3 10 11 15 14 18 19 20 13 12 2 1 546:  5 6 7 17 16 15 11 10 9 8 4 3 2 12 13 14 18 19 20 1 547:  5 6 7 17 16 15 11 12 13 14 18 19 20 1 2 3 10 9 8 4 548:  5 6 7 17 16 15 14 18 19 20 13 12 11 10 9 8 4 3 2 1 549:  5 6 7 17 18 19 20 1 2 3 10 11 12 13 14 15 16 9 8 4 550:  5 6 7 17 18 19 20 13 14 15 16 9 8 4 3 10 11 12 2 1 551:  5 6 19 18 14 13 20 1 2 12 11 15 16 17 7 8 9 10 3 4 552:  5 6 19 18 14 15 11 10 9 16 17 7 8 4 3 2 12 13 20 1 553:  5 6 19 18 14 15 11 12 13 20 1 2 3 10 9 16 17 7 8 4 554:  5 6 19 18 14 15 16 17 7 8 9 10 11 12 13 20 1 2 3 4 555:  5 6 19 18 17 7 8 4 3 2 12 11 10 9 16 15 14 13 20 1 556:  5 6 19 18 17 7 8 9 16 15 14 13 20 1 2 12 11 10 3 4 557:  5 6 19 20 1 2 3 10 9 16 15 11 12 13 14 18 17 7 8 4 558:  5 6 19 20 1 2 12 13 14 18 17 7 8 9 16 15 11 10 3 4 559:  5 6 19 20 13 12 11 10 9 16 15 14 18 17 7 8 4 3 2 1 560:  5 6 19 20 13 14 18 17 7 8 4 3 10 9 16 15 11 12 2 1 5

深搜遍历，递归实现

代码部分：

#include <stdio.h>#include <string.h>int m, f = 1;bool vis[25];int city[25][5];int ans[25];void DFS(int num, int len){int i, t;vis[num] = true;ans[len] = num;for(i = 0; i < 3; i++){t = city[num][i];if(t == m && len == 19){printf("%d:  ", f++);for(int j = 0; j < 20; j++)printf("%d ", ans[j]);printf("%d\n", m);}if(!vis[t])DFS(t, len + 1);}vis[num] = false;}int main(){int i;for(i = 1; i <= 20; i++)scanf("%d%d%d", &city[i][0], &city[i][1], &city[i][2]);while(scanf("%d", &m) && m){memset(vis, false, sizeof(vis));DFS(m, 0);}return 0;}

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