제출 #519862

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
519862		Dasha_Gnedko	Hard route (IZhO17_road)	C++17	19 / 100	2063 ms	24236 KiB

road

#include <bits/stdc++.h>

//#include <ext/pb_ds/detail/standard_policies.hpp>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>

//#pragma GCC optimize("Ofast")
//#pragma GCC target("avx2")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4")

using namespace std;

//using namespace __gnu_pbds;
//template <typename T> using ordered_set = tree <T, null_type, less < T >, rb_tree_tag, tree_order_statistics_node_update>;

mt19937 gen(time(0));

#define ll long long
#define ld long double
#define pb push_back
#define F first
#define S second
#define TIME clock() * 1.0 / CLOCKS_PER_SEC
#define sz(a) int32_t(a.size())
#define endl '\n'
//#define int long long

const int N = 1e6 + 100;
const int M = 31;
const int mod = 1e9 + 7;
const int inf = 1e9 + 7;

vector < int > g[N];
int leaf[N], ra[N], h[N], up[N], pred[N];

void DFS(int v, int pr)
{
    pred[v] = pr;
    h[v] = 0;
    for(auto to: g[v])
    {
        if (to == pr) continue;
        DFS(to, v);
        h[v] = max(h[v], h[to] + 1);
    }
}

void dfs(int v, int pr, int c)
{
    int mx = -2, sr = -2;
    for(auto to: g[v])
    {
        if (to == pr) continue;
        if (h[to] >= mx)
        {
            sr = mx;
            mx = h[to];
        }
        else sr = max(sr, h[to]);
    }
    up[v] = c;
    for(auto to: g[v])
    {
        if (to == pr) continue;
        if (h[to] == mx) dfs(to, v, max(c + 1, sr + 2));
        else dfs(to, v, max(c + 1, mx + 2));
    }
}

ll get_mx(int n)
{
    DFS(0, 0);
    dfs(0, 0, 0);
    ll ans = 0;
    for(int v = 0; v < n; v++)
    {
        vector < int > ve;
        if (up[v]) ve.pb(up[v]);
        for(auto to: g[v])
        {
            if (to == pred[v]) continue;
            ve.pb(h[to] + 1);
        }
        sort(ve.rbegin(), ve.rend());
        while (sz(ve) > 3) ve.pop_back();
        if (sz(ve) == 3)
        {
//            cerr << v + 1 << " " << ve[0] << " " << ve[1] << " " << ve[2] << endl;
            ans = max(ans, 1ll * (ve[1] + ve[2]) * ve[0]);
            ans = max(ans, 1ll * (ve[0] + ve[2]) * ve[1]);
            ans = max(ans, 1ll * (ve[0] + ve[1]) * ve[2]);
        }
    }
    return ans;
}

int calc(int a, int b, int n)
{
    vector < int > pr(n);
    for(int i = 0; i < n; i++)
        ra[i] = -1;
    ra[a] = 0;
    queue < int > q;
    q.push(a);
    while (!q.empty())
    {
        int v = q.front();
        q.pop();
        for(auto to: g[v])
        {
            if (ra[to] == -1)
            {
                ra[to] = ra[v] + 1;
                pr[to] = v;
                q.push(to);
            }
        }
    }
    int mx = ra[b];
    while (1)
    {
        ra[b] = 0;
        q.push(b);
        if (b == a) break;
        b = pr[b];
    }
    for(int i = 0; i < n; i++)
        if (ra[i] != 0) ra[i] = -1;
    while (!q.empty())
    {
        int v = q.front();
        q.pop();
        for(auto to: g[v])
        {
            if (ra[to] == -1)
            {
                ra[to] = ra[v] + 1;
                pr[to] = v;
                q.push(to);
            }
        }
    }
    int cnt = 0;
    for(int i = 0; i < n; i++)
        cnt = max(cnt, ra[i]);
    return cnt * mx;
}

int32_t main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

#ifdef LOCAL
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#else
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);
#endif // LOCAL

    int n;
    cin >> n;
    for(int i = 0; i < n - 1; i++)
    {
        int x, y;
        cin >> x >> y;
        x--; y--;
        g[x].pb(y);
        g[y].pb(x);
    }
    for(int i = 0; i < n; i++)
        leaf[i] = (sz(g[i]) == 1);
    ll mx = get_mx(n), k = 0;
    for(int i = 0; i < n; i++)
    {
        if (!leaf[i]) continue;
        for(int j = i + 1; j < n; j++)
        {
            if (!leaf[j]) continue;
            int c = calc(i, j, n);
            k += (c == mx);
        }
    }
    cout << mx << " " << k;
}

• Subtask 1: 19 / 19
• Subtask 2: 0 / 33
• Subtask 3: 0 / 48