D. Friendly Spiders

By xiaruize

p.s.严厉谴责出题人在题目页面放蜘蛛啊，很难受qwq

思路

暴力建图显然需要 O(n2)O(n^2) , 无法通过本题

考虑是否可以优化, 发现同样的公因子可能被建多条边, 导致边数达到 n2n^2 级别

如何解决这个问题? 可以对于每个公因子建一个点, 将所有有这个因子的数全部与这个点连边, 入边边权为 11 , 出边为 00

进一步优化, 可以发现, 只需要对质因子建点

对于这张图跑 Dijkstra 即可

Code

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/*
    Name:
    Author: xiaruize
    Date:
*/
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define repp(i, x, y) for (int i = x; i >= y; i--)
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 3e5 + 10;

// bool st;
int n;
int a[N];
vector<pii> g[N * 2];
int fa[N * 2];
int siz = 0;
int dis[N * 2];
int st, en;
vector<int> res;
// bool en;

void dfs(int x)//找路径
{
    res.pb(x);
    if (x < N)
        siz++;
    if (x == st)
        return;
    dfs(fa[x]);
}

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    // auto t_1=chrono::high_resolution_clock::now();
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
        int x = a[i];
        for (int j = 2; j * j <= x; j++)
        {
            if (x % j == 0)
            {
                while (x % j == 0)
                    x /= j;
                g[i].pb({j + N, 1});
                g[j + N].pb({i, 0});
            }
        }
        if (x != 1)
        {
            g[i].pb({x + N, 1});
            g[x + N].pb({i, 0});
        }
    }
    cin >> st >> en;
    //Dijkstra
    priority_queue<pii, vector<pii>, greater<pii>> q;
    memset(dis, 0x3f, sizeof(dis));
    dis[st] = 0;
    q.push({0, st});
    while (!q.empty())
    {
        int x = q.top().sec, d = q.top().fir;
        q.pop();
        if (dis[x] < d)
            continue;
        if (x == en)
            break;
        for (auto v : g[x])
        {
            if (dis[v.fir] > dis[x] + v.sec)
            {
                dis[v.fir] = dis[x] + v.sec;
                fa[v.fir] = x;
                q.push({dis[v.fir], v.fir});
            }
        }
    }
    if (dis[en] >= INF)
    {
        cout << "-1" << endl;
        return 0;
    }
    dfs(en);
    reverse(ALL(res));
    cout << siz << endl;
    for (auto v : res)
        if (v < N)
            cout << v << ' ';
    // auto t_2=chrono::high_resolution_clock::now();
    // cout <<". Elapsed (ms): " << chrono::duration_cast<chrono::milliseconds>(t_2 - t_1).count() << endl;
    return 0;
}