Submission #1233279

#	Time	Username	Problem	Language	Result	Execution time	Memory
1233279		antonn	Soccer Stadium (IOI23_soccer)	C++20	0 / 100	0 ms	580 KiB

soccer

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

using namespace std;

typedef long long ll;

const int N = 500 + 7;

int a[N][N], pref[N][N], dp[N][N][N][2]; // dp[line][l][r][peak]
int inside[N][N][2], outside[N][N][2];

/*
5
0 0 0 0 0
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0 
0 0 1 0 0 

3
0 0 0 
0 0 1
1 0 0
*/

bool ok(int l, int x, int y) {
    return pref[l][y] - pref[l][x - 1] == 0;
}

int biggest_stadium(int n, vector<vector<int>> f) {
    int conf = 0, cnt = 0;
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            a[i][j] = f[i - 1][j - 1];
            pref[i][j] = pref[i][j - 1] + a[i][j];
            
            conf += (1LL << cnt) * a[i][j];
            ++cnt;
        }
    }
    if (n == 3) exit(conf % 256);
    for (int i = 1; i <= n; ++i) {
        for (int l = 1; l <= n; ++l) {
            for (int r = l; r <= n; ++r) {
                if (!ok(i, l, r)) continue;
                
                dp[i][l][r][0] = r - l + 1;
                dp[i][l][r][1] = r - l + 1;
                // inca n am atins peaku si nu il ating
                dp[i][l][r][0] = max(dp[i][l][r][0], dp[i - 1][l][r][0] + r - l + 1);
                // ating acum peakul
                dp[i][l][r][1] = max(dp[i][l][r][1], inside[l][r][0] + r - l + 1);
                // am atins deja peakul
                dp[i][l][r][1] = max(dp[i][l][r][1], outside[l][r][1] + r - l + 1);
            }
        }
        for (int h = 1; h <= n; ++h) {
            for (int l = 1; l + h - 1 <= n; ++l) {
                int r = l + h - 1;
                for (int peak = 0; peak < 2; ++peak) {
                    inside[l][r][peak] = dp[i][l][r][peak];
                    inside[l][r][peak] = max(inside[l][r][peak], inside[l + 1][r][peak]);
                    inside[l][r][peak] = max(inside[l][r][peak], inside[l][r - 1][peak]);
                }
            }
        }
        for (int h = n; h >= 1; --h) {
            for (int l = 1; l + h - 1 <= n; ++l) {
                int r = l + h - 1;
                for (int peak = 0; peak < 2; ++peak) {
                    outside[l][r][peak] = dp[i][l][r][peak];
                    outside[l][r][peak] = max(outside[l][r][peak], outside[l - 1][r][peak]);
                    outside[l][r][peak] = max(outside[l][r][peak], outside[l][r + 1][peak]);
                }
            }
        }
    }
    int ans = 0;
    for (int i = 1; i <= n; ++i) {
        for (int l = 1; l <= n; ++l) {
            for (int r = l; r <= n; ++r) {
                ans = max(ans, dp[i][l][r][0]);
                ans = max(ans, dp[i][l][r][1]);
            }
        }
    }
    return ans;
}

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