Codeforces #541 (Div. 2) F. Asya And Kittens (并查集+DFS)

题目 :https://codeforces.com/problemset/problem/1131/F

题目大意: 给你标号为1~N的N个元素,顺序未知,然后N-1次元素合并,每次合并要求两元素所在集合相邻,最后问原始位置.

很明显的看出合并联通块要用到并查集,难点在最后怎么得到原始顺序,看了看巨巨的代码 https://blog.youkuaiyun.com/Link_Ray/article/details/87904632 ,大概就是对每次询问的x,y ,用并查集求其父节点u,v 然后建一条边 u -> v 最后跑一下dfs.因为加边顺序是按输入顺序决定的,所以dfs的时候也是按输入顺序决定的.
比如
6
1 2
1 3
4 5
4 6
1 4
能够得到G[1]={2,3,4} G[4]={5,6} (对应的图也是来自巨巨博客,他给的输入顺序和图好像有点错开,稍微改了下顺序)

当我们对根节点访问的时候,得到的DFS序就刚好满足题意

#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<iostream>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<stdlib.h>
#include<algorithm>
#include<time.h>
#define bug1(g) cout<<"test: "<<g<<endl
#define bug2(g,i) cout<<"test: "<<g<<" "<<i<<endl
#define bug3(g,i,k) cout<<"test: "<<g<<" "<<i<<" "<<k<<endl
using namespace std;
typedef  long long ll;
int n,q;
int x[150005],y[150005],f[150005];
void init()
{
    for(int i =0;i<=n;i++) f[i]=i;
}
int fi(int x)
{
    if(f[x]==x) return x;
    else return f[x]=fi(f[x]);
}
vector<int>a[150005];
void dfs(int x)
{
    cout<<x<<" ";
    //getchar();
    for(int i=0;i<a[x].size();i++)
        dfs(a[x][i]);
}
int main()
{
    ios::sync_with_stdio(0);
    cin>>n;
    init();
    int u,v;
    for(int i =0;i<n-1;i++)
    {
        cin>>x[i]>>y[i];
        u=fi(x[i]),v=fi(y[i]);
        a[u].push_back(v);
        f[v]=u;
    }
    dfs(fi(1));
    return 0;
}