Submission #906128

#	Time	Username	Problem	Language	Result	Execution time	Memory
906128		MilosMilutinovic	Prize (CEOI22_prize)	C++14	19 / 100	3575 ms	397240 KiB

Main

#include <bits/stdc++.h>
using namespace std;
using lint = long long;
using pi = array<int, 2>;
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
const int MAX = 1050000;
 
int n, k, q, t, r1, r2;
int a[MAX], b[MAX], c[MAX], d[MAX];
vector<int> ids;
 
struct tree1 {
	int dep[MAX], par[20][MAX];
	vector<int> gph[MAX];
	vector<pi> adj[MAX];
	lint dist[MAX];
	bool seen[MAX];
	void add(int u, int v) {
		gph[u].push_back(v);
		gph[v].push_back(u);
	}
	void dfs(int x, int px) {
		if (sz(ids) < k) {
			ids.push_back(x);
		}
		dep[x] = dep[px] + 1;
		par[0][x] = px;
		for (int i = 1; i < 20; i++)
			par[i][x] = par[i - 1][par[i - 1][x]];
		for (auto& y : gph[x]) {
			if (y == px) continue;
			dfs(y, x);
		}
	}
	int getlca(int s, int e) {
		if (dep[s] < dep[e]) swap(s, e);
		for (int i = 19; i >= 0; i--) 
			if (dep[par[i][s]] >= dep[e])
				s = par[i][s];
		if (s == e) return s;
		for (int i = 19; i >= 0; i--)
			if (par[i][s] != par[i][e])
				s = par[i][s], e = par[i][e];
		return par[0][s];
	}
	void dfs(int x) {
		seen[x] = true;
		for (auto& y : adj[x]) {
			if (seen[y[0]]) continue;
			dist[y[0]] = dist[x] + y[1];
			dfs(y[0]);
		}
	}
	void build() {
		for (int i = 0; i + 1 < k; i++) {
			int x = ids[i];
			int y = ids[i + 1];
			int z = getlca(ids[i], ids[i + 1]);
			adj[x].push_back({z, -a[i]});
			adj[z].push_back({x, a[i]});
			adj[y].push_back({z, -b[i]});
			adj[z].push_back({y, b[i]});
		}
		dfs(r1);
	}
	lint query(int x, int y) {
		return dist[x] + dist[y] - 2 * dist[getlca(x, y)];
	}
} tree1;
 
struct tree2 {
	int dfn[MAX], T, dep[MAX], par[20][MAX];
	vector<int> gph[MAX];
	vector<pi> adj[MAX];
	lint dist[MAX];
	bool seen[MAX];
	void add(int u, int v) {
		gph[u].push_back(v);
		gph[v].push_back(u);
	}
	void dfs(int x, int px) {
		dfn[x] = ++T;
		dep[x] = dep[px] + 1;
		par[0][x] = px;
		for (int i = 1; i < 20; i++) 
			par[i][x] = par[i - 1][par[i - 1][x]];
		for (auto& y : gph[x]) {
			if (y == px) continue;
			dfs(y, x);
		}
	}
	int getlca(int s, int e) {
		if (dep[s] < dep[e]) swap(s, e);
		for (int i = 19; i >= 0; i--)
			if (dep[par[i][s]] >= dep[e])
				s = par[i][s];
		if (s == e) return s;
		for (int i = 19; i >= 0; i--)
			if (par[i][s] != par[i][e])
				s = par[i][s], e = par[i][e];
		return par[0][s];
	}
	void order() {
		sort(all(ids), [&](int i, int j) { return dfn[i] < dfn[j]; });
	}
	void dfs(int x) {
		seen[x] = true;
		for (auto& y : adj[x]) {
			if (seen[y[0]]) continue;
			dist[y[0]] = dist[x] + y[1];
			dfs(y[0]);
		}
	}
	void build() {
		for (int i = 0; i + 1 < k; i++) {
			int x = ids[i];
			int y = ids[i + 1];
			int z = getlca(ids[i], ids[i + 1]);
			adj[x].push_back({z, -c[i]});
			adj[z].push_back({x, c[i]});
			adj[y].push_back({z, -d[i]});
			adj[z].push_back({y, d[i]});
		}
		vector<int> ord(n);
		iota(all(ord), 0);
		sort(all(ord), [&](int i, int j) { return dfn[i] < dfn[j]; });
		for (int x : ord) if (!seen[x]) dfs(x);
	}
	lint query(int x, int y) {
		return dist[x] + dist[y] - 2 * dist[getlca(x, y)];
	}
} tree2;
 
int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cin >> n >> k >> q >> t;
	for (int i = 0; i < n; i++) {
		int p;
		cin >> p;
		if (p == -1) {
			r1 = i;
		} else {
			--p;
			tree1.add(i, p);
		}
	}
	for (int i = 0; i < n; i++) {
		int p;
		cin >> p;
		if (p == -1) {
			r2 = i;
		} else {
			--p;
			tree2.add(i, p);
		}
	}
	tree1.dfs(r1, r1);
	tree2.dfs(r2, r2);
	tree2.order();
	for (int i = 0; i < k; i++) {
		cout << ids[i] + 1 << " ";
	}
	cout << endl;
	for (int i = 0; i + 1 < k; i++) {
		cout << "? " << ids[i] + 1 << " " << ids[i + 1] + 1 << endl;
	}
	cout << "!" << endl;
	for (int i = 0; i + 1 < k; i++) {
		cin >> a[i] >> b[i] >> c[i] >> d[i];
	}
	tree1.build();
	tree2.build();
	vector<int> x(t), y(t);
	for (int i = 0; i < t; i++) {
		cin >> x[i] >> y[i];
		--x[i]; --y[i];
	}
	for (int i = 0; i < t; i++) {
		cout << tree1.query(x[i], y[i]) << " " << tree2.query(x[i], y[i]) << endl;
	}
}

• Subtask 1: 0 / 10
• Subtask 2: 0 / 25
• Subtask 3: 19 / 19
• Subtask 4: 0 / 22
• Subtask 5: 0 / 24