제출 #848812

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
848812		d4xn	Soccer Stadium (IOI23_soccer)	C++17	25 / 100	325 ms	47700 KiB

soccer

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

using namespace std;

const int inf = INT_MAX;

int n, e;
vector<vector<int>> f;

int biggest_stadium(int N, vector<vector<int>> F) {
    n = N;
    f = F;

    // se pueden coger todas las empty cells?
    bool ok = 1;
    e = 0;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            e += !f[i][j];
        }
    }

    vector<pair<int, int>> s[2];
    s[0].resize(n);
    s[1].resize(n);
    for (int i = 0; i < n; i++) {
        int l = inf;
        int r = -inf;
        for (int j = 0; j < n; j++) {
            if (f[i][j] == 1) continue;
            l = min(l, j);
            r = max(r, j);
        }
        s[0][i] = make_pair(l, r);

        for (int j = l; j <= r; j++) {
            if (f[i][j] == 1) ok = 0;
        }

        l = inf;
        r = -inf;
        for (int j = 0; j < n; j++) {
            if (f[j][i] == 1) continue;
            l = min(l, j);
            r = max(r, j);
        }
        s[1][i] = make_pair(l, r);

        for (int j = l; j <= r; j++) {
            if (f[j][i] == 1) ok = 0;
        }
    }

    for (int x = 0; x < n; x++) {
        for (int y = x+1; y < n; y++) {
            pair<int, int> a = s[0][x];
            pair<int, int> b = s[0][y];
            if (!(a.first <= b.first && b.second <= a.second) && !(b.first <= a.first && a.second <= b.second)) {
                ok = 0;
            }

            a = s[1][x];
            b = s[1][y];    
            if (!(a.first <= b.first && b.second <= a.second) && !(b.first <= a.first && a.second <= b.second)) {
                ok = 0;
            }
        }
    }

    if (ok) return e;

    // hay solo un tree?
    if (e == n*n-1) {
        int ans = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (f[i][j] == 1) {
                    ans = max(ans, n*n - (i+1)*(j-1) + 1);
                    ans = max(ans, n*n - (n-i)*(j+1) + 1);
                    ans = max(ans, n*n - (i+1)*(n-j) + 1);
                    ans = max(ans, n*n - (n-i)*(n-j) + 1);
                    break;
                }
            }
        }
        return ans;
    }

    // else


    

    return 1e9;
}

• Subtask 0: 0 / 0
• Subtask 1: 1.5 / 6
• Subtask 2: 2 / 8
• Subtask 3: 5.5 / 22
• Subtask 4: 4.5 / 18
• Subtask 5: 4 / 16
• Subtask 6: 7.5 / 30