1509: Hamiltonian Cycle

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本文介绍了一个特定的图论问题——汉密尔顿回路问题，并提供了一段C++代码实现。该程序能够接收密集图作为输入，判断是否存在汉密尔顿回路并输出具体的回路序列或无解标识。

1509: Hamiltonian Cycle

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	400	88	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 _{i_l}, v _{i_l}} 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_{i_h} and u_{i_l}

Output

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

or containing:

N

Sample Input

Sample Output

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

This problem is used for contest: 149

#include<iostream>
#include<algorithm>
using namespace std;
int n,map[100][100],vis[100],res[100];
bool flag;
void dfs(int step,int num)
{
    if(flag) return ;
    if(step==n)
    {
        if(map[num][1])
        {
            flag=true;
            printf("%d",res[1]);
            for(int i=2;i<=n;i++) printf(" %d",res[i]);printf(" %d",res[1]);
            printf("/n");
        }
        return ;
    }
    vis[num]=1;
    for(int i=1;i<=n;i++)
    {
        if(map[num][i]&&vis[i]==0)
        {
            res[step+1]=i;
            dfs(step+1,i);
        }
    }
    vis[num]=0;
}
int main()
{
    while(scanf("%d",&n)==1)
    {
        getchar();
        memset(map,0,sizeof(map));
        memset(vis,0,sizeof(vis));
        char str[10];
        while(gets(str))
        {
            if(str[0]=='%') break;
            map[str[0]-'0'][str[2]-'0']=map[str[2]-'0'][str[0]-'0']=1;
        }
        if(n<3) {printf("N/n");continue;}
        flag=false;
        res[1]=1;
        dfs(1,1);
        if(!flag) printf("N/n");
    }
    return 0;
}