HDU 2181 哈密顿绕行世界问题
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
![](https://img2018.cnblogs.com/blog/1435569/201902/1435569-20190205153114121-1175643725.png)
1 #include <iostream> 2 3 using namespace std; 4 5 int Map[21][21] = { 0 }; //邻接矩阵 6 int path[21]; //记录每一次答案的路线 7 bool visited[21] = { false }; 8 int num = 1; //路径的编号 9 10 11 void dfs(int city, int cur) //city为当前访问的城市,已经访问了cur个城市 12 { 13 path[cur] = city; //将当前城市记录在路径上 14 visited[city] = true; 15 16 if (cur == 20) //如果已经访问了20个城市 17 { 18 if (Map[city][path[1]] == 1) //这个城市与起点城市之间存在路径 19 { 20 cout << num << ": "; 21 num++; 22 for (int i = 1; i <= 20; ++i) 23 cout << path[i] << ' '; 24 cout << path[1] << endl; 25 } 26 } 27 28 for (int i = 1; i <= 20; ++i) 29 { 30 if (Map[i][city] == 1 && !visited[i]) 31 { 32 dfs(i, cur+1); //注意此处不能是++cur 33 visited[i] = false; //dfs完要把访问过的顶点重新置为未访问状态 34 } 35 } 36 37 } 38 39 40 int main() 41 { 42 //创建邻接矩阵 43 int city; 44 for (int i = 1; i <= 20; ++i) 45 { 46 for (int j = 1; j <= 3; ++j) 47 { 48 cin >> city; 49 Map[i][city] = 1; 50 } 51 } 52 53 int m; 54 while (cin >> m && m != 0) 55 { 56 dfs(m, 1); 57 num = 1; 58 } 59 60 61 return 0; 62 63 }