joj 1509

 1509: Hamiltonian Cycle

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	467	113	Standard

A few definitions first:

Definition 1

A graph G = (V, E) is called ``dense'' if for each pair of non-adjacent vertices  u and  v,  where  n = | V| and   denotes the degree of the vertex  .

Definition 2

A ``Hamiltonian cycle'' on G is a sequence of vertices (  ) such that  for all   and {  v _il,  v _il} is an edge of G.

The problem is: write a program that, given a dense graph G = (V; E) as input, deter- mines whether G admits a Hamiltonian cycle on G and outputs that cycle, if there is one, or outputs ``N'' if there is none.

Input

A file containing descriptions of graphs, each one ending with a %, in the form:

n₁

%

n₂

%

where n_i is the number of vertices and  are integers between 1 and n indicating that there exists an edge between vertex u_ih and u_il

Output

The output file must contain the sequence of vertices that form a Hamiltonian cycle in the form:

or containing:

N

Sample Input

4
1 2
2 3
2 4
3 4
3 1
%
6
1 2
1 3
1 6
3 2
3 4
5 2
5 4
6 5
6 4
%

Sample Output

1 2 4 3 1
1 2 3 4 5 6 1

This problem is used for contest: 149 

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这是一个哈密尔顿回路问题，其中n必须大于2，因为回路至少要有三个点

#include<cstdio>
#include<string.h>


bool selected;
int map[100][100];
int temp[100];
int store[100];
int n;
void dfs(int x,int y)
{
    if(selected)return ;
    if(y>n&&map[store[1]][store[n]])
    {
        for(int i=1;i<=n;i++)printf("%d ",store[i]);
        printf("1\n");
        selected=true;
        return ;
    }
    for(int i=1;i<=n;i++)
    {
        if(map[x][i]&&!temp[i])
        {
            temp[i]=1;
            store[y]=i;
            dfs(i,y+1);
            temp[i]=0;
        }
    }
}
int main()
{
    while(scanf("%d",&n)==1)
    {
        getchar();
        memset(map,0,sizeof(map));
        memset(temp,0,sizeof(temp));
        char str[4];
        gets(str);
        while(strcmp(str,"%")!=0)
        {
            int a = str[0] - '0';
            int b = str[2] - '0';
            map[a][b] = map[b][a] = 1;
            gets(str);
        }
        if(n<3)
        {
            printf("N\n");
            continue;
        }
        else//此时的else必须与if连在一起因为n<3时候也必须要有输入！！！
        {
            selected=false;
            store[1]=1;
            temp[1]=1;
            dfs(1,2);
            if(!selected)printf("N\n");
        }
    }
    return 0;
}