漢密爾頓圖：

定義：哈密頓通路（迴路）與哈密頓圖 （Hamilton圖） 通過圖G的每個結點一次，且僅一次的通路（迴路），就是哈密頓通路（迴路）。存在哈密頓迴路的圖就是哈密頓圖

Granny's Bike

釋出時間: 2017年6月19日 00:27   時間限制: 1000ms   記憶體限制: 128M

Most days Granny rides her bike around town to do errands, visit, have a cup of coffee, and so on. She enjoys riding her bike and wants to avoid passing the same place twice to add to the interest of the ride. So, each day she draws a map of the places to be visited, with lines connecting those near each other, and sees if she can visit them all and return home without passing a place more than once. Some days she finds she can do this and other days she finds she can't. For example, for the map on the left, Granny can visit every place and return home without passing any place twice, but she can't do it for the map on the right.

[圖片]

She turns to you to write a program to help her.

There will be multiple test cases for this problem. Each test case will have input on multiple lines. The first line will contain the integer n (< 10) noting the number of places Granny wants to visit that day. These will be numbered 1 through n and Granny's house will be numbered 0. The next n lines will be a list of those places near each spot. The first line will be a list of places with a direct route from place 1. The second line will be a list of places with a direct route from place 2, and so on. You may assume that if place i has a direct route to place j, then there is a direct route the other direction also. A line containing 0 will follow the last test case.

For each test case, print one line of the form:Case m: Granny can make the circuit.
Or Case m: Granny can not make the circuit as appropriate. Here, m is the number of the test case, starting at 1.

5
0 2 5
0 1 3
2 4
0 3 5
1 4
4
0 2 3 4
1 3
1 2
0 1
0

Case 1: Granny can make the circuit.
Case 2: Granny can not make the circuit.

題意：從0點出發，輸入資料n代表要去旅行的城市數目，之後n行表示第i行即第i個城市與哪幾個城市相連，求能否每個城市都旅行一次再返回起始點0點

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
typedef long long ll;
typedef long long LL;
int tu[25][25];
int vis[25];
int n;
int sum;
int flag;
bool dfs(int k)
{
    if(sum==n+1&&k==0)
    {
        return true;
    }

    for(int i=0; i<=n; i++)
    {
        if(!vis[i]&&tu[k][i])
        {
            sum++;
            vis[i]=1;
            if(dfs(i))
            {
                return true;
            }
            sum--;
            vis[i]=0;
        }
    }
    return false;
}
int main()
{
    char a[25];
    int hh;
    int len;
    int t=0;
    while(scanf("%d",&n)&&n)
    {
        flag=0;
        sum=0;
        memset(tu,0,sizeof(tu));
        memset(vis,0,sizeof(vis));
        getchar();
        for(int i=1; i<=n; i++)
        {
            memset(a,0,sizeof(a));
            gets(a);
            len=strlen(a);
            for(int j=0; j<len; j+=2)
            {
                hh=a[j]-'0';
                tu[i][hh]=tu[hh][i]=1;
            }
        }
        if(dfs(0))
        {
            cout<<"Case "<<++t<<":"<<" Granny can make the circuit."<<endl;
        }
        else
        {
            cout<<"Case "<<++t<<":"<<" Granny can not make the circuit."<<endl;
        }
    }
    return 0;
}