Submission #1158270

#	Time	Username	Problem	Language	Result	Execution time	Memory
1158270		countless	September (APIO24_september)	C++20	0 / 100	0 ms	324 KiB

september

#include "september.h"
#include <bits/stdc++.h>
#include <vector>

using namespace std;

int dfs(int node, vector<vector<int>> &down, vector<bool> &vis) {
	int res = 0;

	for (auto child : down[node]) {
		if (!vis[child]) {
			vis[child] = true;
			res++;
			res += dfs(child, down, vis);
		}
	}

	return res;
}

int solve(int N, int M, std::vector<int> F, std::vector<std::vector<int>> S) {
	int ans = 0;
	vector<vector<int>> down(N, vector<int>());
	
	for (int i = 1; i < N - 1; i++) down[F[i]].push_back(i);

	int dels = 0;
	int sweep = 0;
	vector<bool> vis(N, false);
	vector<int> cnt(N, 0);

	for (int i = 0; i < N; i++) {
		for (int m = 0; m < M; m++) {
			// delete itself and all its children
			// not including vis'd
			if (!vis[S[m][i]]) {
				vis[S[m][i]] = true;
				dels++;

				dels += dfs(S[m][i], down, vis);
			}

			// increment how many times we've seen this node
			cnt[S[m][i]]++;

			// interval start and end
			// first time we've seen it
			if (cnt[S[m][i]] == 1) sweep++;
			// last time we've seen it
			if (cnt[S[m][i]] == M) sweep--;
		}

		// when an interval "closes," increment ans
		// an interval closes when
		// we've seen all the nodes in the interval
		// specifically, for any node, it's subtree is contained in the interval
		// thus, this is only possible when the amount of deletions
		// is equal to the amount of nodes in the interval
		if (i + 1 == dels && sweep == 0) {
			ans++;
		}
	}

	return ans;
}

• Subtask 1: 0 / 11
• Subtask 2: 0 / 14
• Subtask 3: 0 / 5
• Subtask 4: 0 / 9
• Subtask 5: 0 / 5
• Subtask 6: 0 / 11
• Subtask 7: 0 / 9
• Subtask 8: 0 / 11
• Subtask 9: 0 / 9
• Subtask 10: 0 / 16