生成一课n个节点的树。
n表示点数，通过并查集判断无环

#include <ctime>
#include <iostream>
#include<random>

using namespace std;

const int N = 1e6 + 10, INF = 1e9;
int a[N];
int p[N];
vector<int> edge[N];

int find(int x)
{
    if(p[x] != x)
        return p[x] = find(p[x]);
    return p[x];
}


int main() {
    mt19937 rnd(time(0));
    freopen("5.in","w",stdout);
    int n= 1000;
    for(int i = 1; i <= n; i ++)
        p[i] = i;
    printf("%d\n", n);
    for(int i=1;i <= n; i ++){
        a[i]= rnd()% INF + 1;
        printf("%d ", a[i]);
    }
    printf("\n");
    int cnt = 0;
    while(cnt < n - 1)
    {
        int u = rnd() % n + 1;
        int v = rnd() % n + 1;
        int fu = find(u);
        int fv = find(v);
        if(fu != fv)
        {
            cnt ++;
            p[fu] = fv;
            edge[u].push_back(v);
            edge[v].push_back(u);
            printf("%d %d\n", u, v);
        }
    }

}