C++ :

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
bool a[110][110];
bool flag[110];
int n,e;
void init();
void dfs(int x);
int main()
{
	//freopen("dfs1.in","r",stdin);
	//freopen("dfs5.out","w",stdout);
	init();
	dfs(1);
	return 0;
}
void init()
{
	cin>>n>>e;
	memset(flag,0,sizeof(flag));
	memset(a,0,sizeof(a));
	int x,y;
	for(int i=1;i<=e;++i)
	{
		cin>>x>>y;
		a[x][y]=1;//无向图处理双向 
		a[y][x]=1;
	}
}
void dfs(int x)
{
	cout<<x<<' ';
	flag[x]=1;//把遍历点设为已访问 
	for(int i=1;i<=n;++i) //寻找相邻连通的下一个节点 
	  if(!flag[i]&&a[x][i]==1) dfs(i);//如果找到以这个节点接着遍历 
}

Pascal :

program acm21182;
const maxn=100;
type node=record
          u,v,next:longint;
          end;
var
  num,i,j,k,n,e,u,v,t:longint;
  graph:array[1..maxn,1..maxn] of longint;
  vis:array[1..maxn] of boolean;
procedure dfs(x:longint);
var
  i:longint;
begin
  if k=0 then
   begin
     write(x); k:=k+1; 
   end
  else write(' ',x);
  vis[x]:=true;
   for i:=1 to n do
     if (not vis[i]) and (graph[x,i]>0) then
       begin
         graph[x,i]:=0;
         graph[i,x]:=0;
         dfs(i);
       end;
end;
procedure init;
var i:longint;
begin
  fillchar(graph,sizeof(graph),0);
  readln(n,e);
  for i:=1 to e do
   begin
     readln(u,v);
     graph[u,v]:=1;
     graph[v,u]:=1;
   end;
end;
begin
  fillchar(vis,sizeof(vis),0);
  init;
  k:=0;
  dfs(1);
end.