HDU 2181哈密顿绕行世界问题(还是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): 2670    Accepted Submission(s): 1673

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

原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=2181

水题,和HDU1016差不多,那题的题解:http://blog.csdn.net/hurmishine/article/details/50926454

AC代码1;

#include <iostream>
#include <cstring>
using namespace std;
int ans[25];
bool a[25][25];
bool vis[25];
int start;
int kase;
void DFS(int now,int num)//2
{
    if(num==21&&a[start][now])
    {

        cout<<++kase<<":  ";
        for(int i=1; i<=20; i++)
            cout<<ans[i]<<" ";
        cout<<start<<endl;
        return ;
    }
    for(int i=1; i<=20; i++)
    {
        if(!vis[i]&&a[now][i])
        {
            vis[i]=true;
            ans[num]=i;
            DFS(i,num+1);
            vis[i]=false;
        }
    }
}
int main()
{
    int k;
    memset(a,false,sizeof(a));
    for(int i=1; i<=20; i++)
    {
        for(int j=0; j<3; j++)
        {
            cin>>k;
            a[i][k]=a[k][i]=true;
        }
    }
    while(cin>>start)
    {
        if(start==0) break;
        memset(vis,false,sizeof(vis));
        vis[start]=true;
        ans[1]=start;
        kase=0;
        DFS(start,2);
    }
    return 0;
}

AC代码2:

#include <iostream>
#include <cstring>
using namespace std;
int ans[25];
bool a[25][25];
bool vis[25];
int start;
int kase;
void DFS(int num)//2
{
    if(num==21&&a[start][ans[20]])
    {

        cout<<++kase<<":  ";
        for(int i=1; i<=20; i++)
            cout<<ans[i]<<" ";
        cout<<start<<endl;
        return ;
    }
    for(int i=1; i<=20; i++)
    {
        if(!vis[i]&&a[ans[num-1]][i])
        {
            vis[i]=true;
            ans[num]=i;
            DFS(num+1);
            vis[i]=false;
        }
    }
}
int main()
{
    int k;
    memset(a,false,sizeof(a));
    for(int i=1; i<=20; i++)
    {
        for(int j=0; j<3; j++)
        {
            cin>>k;
            a[i][k]=a[k][i]=true;
        }
    }
    while(cin>>start)
    {
        if(start==0) break;
        memset(vis,false,sizeof(vis));
        vis[start]=true;
        ans[1]=start;
        kase=0;
        DFS(2);
    }
    return 0;
}