C

Think:
1DFS:
1>控制结束条件
2>传递
2反思:
1>不自信
2>未判断最后一个城市是否可以到达初始出发城市（感谢鑫哥点拨）
3收获:
1>DFS加深理解
2>通过宏观分步调试判断出现错误地方
3>哈密顿图理解:不重复经过每个结点回到最初出发结点

vjudge题目链接

以下为Accepted代码

#include <cstdio>#include <cstring>using namespace std;struct node{    int t[4];    int front;} a[24];int m, k, cnt, v[24];void Input();void DFS(int x);int main(){    Input();    while(scanf("%d", &m) && m)    {        k = cnt = 0;        memset(v, 0, sizeof(v));        a[m].front = -1;        v[m] = 1, cnt++;        DFS(m);    }    return 0;}void DFS(int x){    int i;    if(cnt == 20)    {        int flag = 0;        for(i = 1; i <= 3; i++){            if(a[x].t[i] == m){                flag = 1;                break;            }        }        if(!flag)            return;        int tp, link[24];        tp = 0;        link[tp++] = m;        link[tp++] = x;        int c = a[x].front;        while(c != -1)        {            link[tp++] = c;            c = a[c].front;        }        k++;        printf("%d: ", k);        for(i = tp-1; i >= 0; i--)            printf(" %d", link[i]);        printf("\n");    }    else    {        for(i = 1; i <= 3; i++)        {            if(v[a[x].t[i]] == 0)            {                cnt++;                a[a[x].t[i]].front = x;                v[a[x].t[i]] = 1;                DFS(a[x].t[i]);                cnt--;                v[a[x].t[i]] = 0;            }        }    }}void Input(){    for(int i = 1; i <= 20; i++)    {        for(int j = 1; j <= 3; j++)        {            scanf("%d", &a[i].t[j]);        }    }}