이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#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;
}
}
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