• HDU2181:哈密顿绕行世界问题(DFS)


    哈密顿绕行世界问题

    Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)
    Total Submission(s) : 57   Accepted Submission(s) : 31

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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<set>
    #include<cstring>
    using namespace std;
    int ans,i,j,a,b,c,m;
    int f[25],vis[25];
    set<int>s[22];
    void dfs(int k,int l)
    {
        if(l==21)
        {
            if (f[l]==m)
            {
                ans++;
                printf("%d: ",ans);
                for(int i=1;i<=21;i++)
                printf(" %d",f[i]);
                printf("\n");
            }
            return;
        }
        set<int>::iterator it=s[k].begin();
        while(it!=s[k].end())
        {
            if (vis[*it]){++it; continue;}
            vis[*it]=1;
            f[l+1]=*it;
            dfs(f[l+1],l+1);
            vis[*it]=0;
            it++;
        }
        return;
    }
    int main()
    {
        for(i=1;i<=20;i++)
            {
                scanf("%d%d%d",&a,&b,&c);
                s[i].insert(a);
                s[i].insert(b);
                s[i].insert(c);
            }
        while(scanf("%d",&m),m)
        {
            memset(vis,0,sizeof(vis));
           // vis[m]=1;
            f[1]=m;
            ans=0;
            dfs(m,1);
    
        }
        return 0;
    }
    #include <iostream>
    #include<cstdio>
    #include<set>
    #include<cstring>
    using namespace std;
    int ans,i,j,a,b,c,m;
    int f[25],vis[25];
    set<int>s[22];
    void dfs(int k,int l)
    {
        if(l==21)
        {
            if (f[l]==m)
            {
                ans++;
                printf("%d: ",ans);
                for(int i=1;i<=21;i++)
                printf(" %d",f[i]);
                printf("\n");
            }
            return;
        }
        set<int>::iterator it=s[k].begin();
        while(it!=s[k].end())
        {
            if (vis[*it]){++it; continue;}
            vis[*it]=1;
            f[l+1]=*it;
            dfs(f[l+1],l+1);
            vis[*it]=0;
            it++;
        }
        return;
    }
    int main()
    {
        for(i=1;i<=20;i++)
            {
                scanf("%d%d%d",&a,&b,&c);
                s[i].insert(a);
                s[i].insert(b);
                s[i].insert(c);
            }
        while(scanf("%d",&m),m)
        {
            memset(vis,0,sizeof(vis));
           // vis[m]=1;
            f[1]=m;
            ans=0;
            dfs(m,1);
    
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/stepping/p/5669085.html
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