F. Asya And Kittens

 http://codeforces.com/contest/1131/problem/F

题意：有n只猫相邻有隔板隔着，n-1对(x,y)输入表示取下某一隔板能使x与y猫在一起，取下后两笼合并，里面的猫都与相邻笼的猫相邻，输出一组序列，能符合顺序地使得猫在一起；

思路：并查集；连接每个点集合；

#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<sstream>
#include<cstdio>
#include<ctime>
#include<map>
#include<stack>
#include<string>
using namespace std;

#define sfi(i) scanf("%d",&i)
#define pri(i) printf("%d\n",i)
#define sff(i) scanf("%lf",&i)
#define ll long long
#define mem(x,y) memset(x,y,sizeof(x))
#define INF 0x3f3f3f3f
#define eps 1e-6
#define PI acos(-1)
#define lowbit(x) ((x)&(-x))
#define zero(x) (((x)>0?(x):-(x))<eps)
#define fl() printf("flag\n")
ll gcd(ll a,ll b){while(b^=a^=b^=a%=b);return a;}
const int maxn=3e5+9;
const int mod=1e9+7;

template <class T>
inline void sc(T &ret)
{
    char c;
    ret = 0;
    while ((c = getchar()) < '0' || c > '9');
    while (c >= '0' && c <= '9')
    {
        ret = ret * 10 + (c - '0'), c = getchar();
    }
}

vector<int>g[maxn];
int p[maxn];
void dfs(int u,int fa)
{
    cout<<u<<" ";
    for(int i=0;i<g[u].size();i++)
    {
        int v=g[u][i];
        if(v!=fa)
        {
            dfs(v,u);
        }
    }
}
int Find(int x)
{
    return p[x]==x?x:p[x]=Find(p[x]);
}
void join(int x,int y)
{
    int jx=Find(x);
    int jy=Find(y);
    if(g[jx].size()<g[jy].size()) swap(jx,jy);
    //cout<<jx<<" "<<jy<<endl;
    if(jx!=jy)
    {
        for(int i=0;i<g[jy].size();i++)
        {
            g[jx].push_back(g[jy][i]);
            p[g[jy][i]]=jx;
        }
    }


}
int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++) p[i]=i,g[i].push_back(i);
    for(int i=0;i<n-1;i++)
    {
        int u,v;
        cin>>u>>v;
        join(u,v);
    }
    for(int i=1;i<=n;i++)
    {
        if(p[i]==i)
        {
            for(int j=0;j<g[i].size();j++)
            {
                cout<<g[i][j]<<" ";
            }
            break;
        }
    }
    return 0;
}