Ehab's REAL Number Theory Problem CodeForces

参考文章：https://oierwanhong.cc/2020/03/16/CF1325E/
每个数至多有个7约数，也就是每个数至多两个不同的质因数
如果某个数没有指数为奇数的质因子，那么它就是一个完全平方数，直接输出1
如果某个数只有1个指数为奇数的质因子，让这个质因子和"源点"1连无向边
某个数有两个指数为奇数的质因子，让这两个质因子间连无向边
然后分别bfs每个点，去求最小环

#include<set>
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define f first
#define s second
#define pii pair<int,int>
#define int long long
int read()
{
    int res=0,ch,flag=0;
    if((ch=getchar())=='-')             //判断正负
        flag=1;
    else if(ch>='0'&&ch<='9')           //得到完整的数
        res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')
        res=res*10+ch-'0';
    return flag?-res:res;
}
const int N=1e6+7;
set<pii>ms;
int e[N],ne[N],h[N],idx;
void add(int a,int b) {
	e[idx]=b;
	ne[idx]=h[a];
	h[a]=idx++;
}
void add_edge(int u,int v) {
	if(v<u)
		swap(u,v);
	if(ms.find({u,v})!=ms.end()) {
		cout<<2<<endl;
		exit(0);
	}
	ms.insert({u,v});
	add(u,v);
	add(v,u);
}
int dist[N],son[N];
int ans=0x3f3f3f3f;
void bfs(int S) {
	memset(dist,0x3f,sizeof dist);
	queue <int> q;
	q.push(S);
	dist[S] = 0;
	vector <int> who;
	while (q.size()) {
		int u = q.front();
		q.pop();
		who.push_back(u);
		for (int i=h[u]; i!=-1; i=ne[i]) {
			int v=e[i];
			if (dist[u] + 1 < dist[v]) {
				dist[v] = dist[u] + 1;
				if (u == S)
					son[v] = v;
				else
					son[v] = son[u];
				q.push(v);
			} else if (u != S && v != S && son[u]!=son[v])
				ans=min(ans,dist[u]+dist[v]+1);
		}
	}
}
signed main() {
	memset(h,-1,sizeof h);
	int n=read();
	vector<pii>ed;
	while(n--) {
		int x=read();
		vector<int>a;
		for(int i=2; i*i<=x; i++) {
			if(x%i==0) {
				int p=0;
				while(x%i==0) {
					++p;
					x/=i;
				}
				if(p&1)
					a.push_back(i);
			}
		}
		if(x>1)
			a.push_back(x);
		if(a.empty()) {
			cout<<1<<endl;
			exit(0);
		} else if(a.size()==1)
			ed.push_back({1,a[0]});
		else if(a.size()==2)
			ed.push_back({a[0],a[1]});
	}
	for(auto e:ed)
		add_edge(e.f,e.s);
	for(int u=1; u<=1007; u++)
		bfs(u);
	if(ans>=0x3f3f3f3f)
		cout<<"-1"<<endl;
	else
		cout<<ans<<endl;
	return 0;
}