D. Friendly Spiders（bfs最短路径、质数虚点建图）

题意：

• 给一个长度为 n 的数组 a ，数组中两两不互质的数可以建一条边，即$gcd(a[i],a[j]) ≠ 1$，i，j之间存在伊奥无向边
• 问 s 到 t 的最短路径是多长，并输出

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题解

• 根据唯一分解定理，所有的数都可以表示为$\prod p_i^{k_j}$
• 1e9之内所有的数，其不同质数数量不会很多（本体数据3e5，更少了，不会超过10个）
• 所以我们可以利用这些质数来建边

---

#include<bits/stdc++.h>
#define debug1(a) cout<<#a<<'='<< a << endl;
#define debug2(a,b) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<endl;
#define debug3(a,b,c) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<"  "<<#c<<" = "<<c<<endl;
#define debug4(a,b,c,d) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<"  "<<#c<<" = "<<c<<"  "<<#d<<" = "<<d<<endl;
#define debug5(a,b,c,d,e) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<"  "<<#c<<" = "<<c<<"  "<<#d<<" = "<<d<<"  "<<#e<<" = "<<e<<endl;
#define endl "\n"
#define fi first
#define se second

//#define int long long
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef pair<LL,LL> PLL;

//#pragma GCC optimize(3,"Ofast","inline")
//#pragma GCC optimize(2)
const int N = 3e5+10;
int a[N];
int n,s,t;

//int primes[N], cnt;
bool st[N];
vector<vector<int>> pr(N+1),edge(2*N+1);
int dist[2*N];
int pre[2*N];

void get_primes(int n)
{
    for (int i = 2; i <= n; i ++ )
    {
        if (st[i]) continue;
        // primes[cnt ++ ] = i;
        for (int j = i; j <= n; j += i){
            st[j] = true;
            pr[j].push_back(i);
        }
    }
}

void solve() 
{
    memset(dist,0x3f,sizeof dist);

    cin >> n;
    for(int i = 1;i <= n;i ++)cin >> a[i];
    cin >> s >> t;

    get_primes(N-1);

    for(int i = 1;i <= n;i ++)
    {
        for(auto x:pr[a[i]])
        {
            //debug2(i,x+n);
            edge[i].push_back(x + n);
            edge[x + n].push_back(i);
        }
    }

    dist[s] = 0;
    queue<int> q;q.push(s);
    while(q.size())
    {
        int y = q.front();q.pop();
        for(auto x:edge[y])
        {
            //debug2(y,x);
            if(dist[x] > dist[y] + 1)
            {
                dist[x] = dist[y] + 1;
                pre[x] = y;
                q.push(x);
            }
        }
    }

    if(dist[t] == 0x3f3f3f3f)
    {
        cout << -1 << endl;
        return ;
    }

    stack<int> sta;sta.push(t);
    while(t != s)
    {
        t = pre[t];
        if(t <= n)sta.push(t);
    }
    cout << sta.size() << endl;
    while(sta.size())
    {
        cout << sta.top() << " ";
        sta.pop();
    }

}

signed main()
{
    
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
    int T = 1;//cin >> T;

    while(T--){
        //puts(solve()?"YES":"NO");
        solve();
    }
    return 0;

}
/*

*/