codeforces #541 F Asya And Kittens(并查集+输出路径)

F. Asya And Kittens

Asya loves animals very much. Recently, she purchased n−1 partitions originally. Initially, each cell contained exactly one kitten with some number.

Observing the kittens, Asya noticed, that they are very friendly and often a pair of kittens in neighboring cells wants to play together. So Asya started to remove partitions between neighboring cells. In particular, on the day i, Asya:

• Noticed, that the kittens yi, located in neighboring cells want to play together.
• Removed the partition between these two cells, efficiently creating a single cell, having all kittens from two original cells.

Since Asya has never putted partitions back, after n−1 days the cage contained a single cell, having all kittens.

For every day, Asya remembers numbers of kittens n cells.

Input

The first line contains a single integer 2≤n≤150000) — the number of kittens.

Each of the following xi≠yi) — indices of kittens, which got together due to the border removal on the corresponding day.

It's guaranteed, that the kittens yi were in the different cells before this day.

Output

For every cell from n, who was originally in it.

All printed integers must be distinct.

It's guaranteed, that there is at least one answer possible. In case there are multiple possible answers, print any of them.

Example

input

5

1 4

2 5 

4 5

output

3 1 4 2 5

 

题意：模拟合并两个连通块

思路：用链表的思想去合并

#include <cstdio>
#include <map>
#include <iostream>
#include<cstring>
#include<bits/stdc++.h>
#define ll long long int
#define M 6
using namespace std;
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
int moth[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int dir[4][2]={1,0 ,0,1 ,-1,0 ,0,-1};
int dirs[8][2]={1,0 ,0,1 ,-1,0 ,0,-1, -1,-1 ,-1,1 ,1,-1 ,1,1};
const int inf=0x3f3f3f3f;
const ll mod=1e9+7;
int f[200000],n;
int path[200000],last[200000]; //path记录路径 last记录最后一个数 
int find(int x){
    if(x!=f[x])
    f[x]=find(f[x]);
    return f[x];
} 
void join(int x,int y){
    int xx=find(x);
    int yy=find(y);
    if(xx!=yy)
        f[yy]=xx;
}
int main(){
    ios::sync_with_stdio(false);
    cin>>n;
    for(int i=1;i<=n;i++) f[i]=i,last[i]=i;
    n--;
    while(n--){
        int x,y; cin>>x>>y;
        int xx=find(x); int yy=find(y);
        if(xx==yy) continue;
        path[last[xx]]=yy;
        last[xx]=last[yy];
        join(x,y);
    }
    for(int i=find(1);i;i=path[i])
    cout<<i<<" ";
    return 0;
}