Submission #955202

#	Time	Username	Problem	Language	Result	Execution time	Memory
955202		42kangaroo	Soccer Stadium (IOI23_soccer)	C++17	12 / 100	3602 ms	2097152 KiB

soccer

#include "soccer.h"
#include "bits/stdc++.h"


int biggest_stadium(int N, std::vector<std::vector<int>> F) {
	using namespace std;
	using dp_t = vector<vector<vector<vector<int>>>>;
	bool allPos = true;
	vector<pair<int, int>> ra(N);
	vector<int> les(N);
	for (int i = 0; i < N; ++i) {
		ra[i].first = 0;
		for (; ra[i].first < N; ra[i].first++) {
			if (F[i][ra[i].first] == 0) break;
		}
		ra[i].second = N - 1;
		for (; ra[i].second >= 0; ra[i].second--) {
			if (F[i][ra[i].second] == 0) break;
		}
		les[i] = ra[i].second - ra[i].first + 1;
		for (int j = ra[i].first; j <= ra[i].second; ++j) {
			if (F[i][j] == 1) allPos = false;
		}
		if (!allPos) break;
	}
	int tot = 0;
	int maI = max_element(les.begin(), les.end()) - les.begin();
	pair<int,int> actRa = ra[maI];
	pair<int,int> oRa = {maI, maI};
	tot = les[maI];
	while (allPos && oRa != make_pair(0, N - 1)) {
		int ne = oRa.first -1;
		if (ne < 0 || (oRa.second < N - 1 && les[ne] < les[oRa.second + 1])) ne = oRa.second + 1;
		if (les[ne] < 0) break;
		if (actRa.first > ra[ne].first || actRa.second < ra[ne].second) allPos = false;
		actRa = ra[ne];
		tot += les[ne];
		oRa = {min(oRa.first, ne), max(oRa.second, ne)};
	}
	if (allPos) return tot;
	else {
		vector<vector<int>> prefS(N, vector<int>(N + 1, 0));
		for (int i = 0; i < N; ++i) {
			for (int j = 0; j < N; ++j) {
				prefS[i][j + 1] = prefS[i][j] + F[i][j];
			}
		}
		dp_t dp(N, vector<vector<vector<int>>>(N, vector<vector<int>>(N, vector<int>(N, -1))));
		int ma = 0;
		for (int i = 0; i < N; ++i) {
			for (int j = i; j < N; ++j) {
				for (int k = 0; k < N; ++k) {
					for (int l = N - 1; l >= k; --l) {
						if (l < N - 1) dp[i][j][k][l] = dp[i][j][k][l + 1];
						if (k > 0) dp[i][j][k][l] = max(dp[i][j][k][l], dp[i][j][k - 1][l]);
						if (i == j) {
							if (prefS[i][l + 1] - prefS[i][k] == 0) dp[i][j][k][l] = max(dp[i][j][k][l], l - k + 1);
						} else {
							if (prefS[i][l + 1] - prefS[i][k] == 0) dp[i][j][k][l] = max(dp[i][j][k][l], l - k + 1 + dp[i + 1][j][k][l]);
							if (prefS[j][l + 1] - prefS[j][k] == 0) dp[i][j][k][l] = max(dp[i][j][k][l], l - k + 1+ dp[i][j - 1][k][l]);
						}
						ma = max(ma, dp[i][j][k][l]);
					}
				}
			}
		}
		return ma;
	}
}

• Subtask 0: 0 / 0
• Subtask 1: 0 / 6
• Subtask 2: 2 / 8
• Subtask 3: 5.5 / 22
• Subtask 4: 4.5 / 18
• Subtask 5: 0 / 16
• Subtask 6: 0 / 30