Submission #966309

#	Time	Username	Problem	Language	Result	Execution time	Memory
966309		eysbutno	Hard route (IZhO17_road)	C++17	100 / 100	608 ms	140388 KiB

road

/**
 * Assume there is some hard route that goes from vertex
 * u to vertex v. Let the node that the path from u to v and the
 * furthest node from the hard route meet be node x. Use rerooting
 * DP to calculate the hardest route and the # of such hardest routes
 * for every node x.
 * 
 * Time Complexity: O(n log(n))
*/
#include <bits/stdc++.h>
using namespace std;
using ll = long long;

int main() {
    cin.tie(0) -> sync_with_stdio(0);
    int n; 
    cin >> n;
    vector<vector<int>> adj(n);
    for (int i = 1; i < n; i++) {
        int x, y; 
        cin >> x >> y;
        --x, --y;
        adj[x].push_back(y);
        adj[y].push_back(x);
    }

    vector<int> max_length(n), path_count(n);
    auto dfs = [&](int u, int p, auto&& dfs) -> void {
		/**
         * Calculates the longest path from vertex u,
         * and the number of such paths.
        */
        max_length[u] = 0;
        path_count[u] = 1;
        for (int v : adj[u]) if (v != p) {
            dfs(v, u, dfs);
            if (max_length[u] < max_length[v] + 1) {
                max_length[u] = max_length[v] + 1;
                path_count[u] = path_count[v];
            } else if (max_length[v] + 1 == max_length[u]) {
                path_count[u] += path_count[v];
            }
        }
    }; 
    dfs(0, -1, dfs);

    ll hard = 0, cnt = 1;
    auto dfs2 = [&](int u, int p, ll parDist, ll parCnt, 
                    auto&& dfs2) -> void {
		/**
         * Performs the rerooting, to count the hardest
         * path and the # of such paths at this vertex.
        */
        vector<array<ll, 2>> paths; // {distance, count}
        if (u > 0 || (int) adj[u].size() == 1) {
            paths.push_back({parDist, parCnt});
        }
        for (int v : adj[u]) if (v != p) {
            paths.push_back({max_length[v] + 1, path_count[v]});
        }
        sort(paths.begin(), paths.end(), greater<>());
        if ((int) adj[u].size() >= 3) { // can form a nonzero hard route
            /**
             * Let the 3 longest path lengths be a, b, c, with a > b > c.
             * The optimal hard route "hardness" is a * (b + c).
            */
            ll a = paths[0][0], b = paths[1][0], c = paths[2][0];
            ll cur = a * (b + c), num = 0, ties = 0;
            for (auto [k, v] : paths) {
                if (k == c) ties += v;
            }

            // case 1: all are distinct.
            if (a != b && b != c) {
                num = paths[1][1] * ties;
            }
            // case 2: all are the same.
            else if (a == b && b == c) {
                num = ties * ties;
				for (auto [k, v] : paths) {
					if (k == a) num -= v * v;
				}
				num /= 2; // avoiding double counting
            }
            // case 3: first two are the same.
            else if (a == b) {
                num = (paths[0][1] + paths[1][1]) * ties;
            }
            // case 4: last two are the same.
            else {
               	num = ties * ties;
				for (auto [k, v] : paths) {
					if (k == c) num -= v * v;
				}
				num /= 2;
            }
            if (hard < cur) {
                hard = cur, cnt = num;
            } else if (hard == cur) {
                cnt += num;
            }
        }
        // processing parent dist and parent count.
        ll l1 = 0, l2 = 0, cnt1 = 0, cnt2 = 0;
        for (auto [k, v] : paths) {
            if (k + 1 > l1) {
                swap(l1, l2);
                swap(cnt1, cnt2);
                l1 = k + 1, cnt1 = v;
            } else if (k + 1 == l1) {
                cnt1 += v;
            } else if (k + 1 > l2) {
                l2 = k + 1, cnt2 = v;
            } else if (k + 1 == l2) {
                cnt2 += v;
            }
        }
        for (int v : adj[u]) if (v != p) {
            // using the best parent hardness and parent count possible.
            if (max_length[v] + 2 == l1) {
                (path_count[v] == cnt1) ? dfs2(v, u, l2, cnt2, dfs2) :
                            	          dfs2(v, u, l1, cnt1 - path_count[v], dfs2);
            } else {
                dfs2(v, u, l1, cnt1, dfs2);
            }
        }
    }; dfs2(0, -1, 0, 1, dfs2);
    cout << hard << ' ' << cnt << '\n';
}

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