제출 #839800

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
839800		mircea_007	축구 경기장 (IOI23_soccer)	C++17	0 / 100	1 ms	336 KiB

soccer

#include <stdio.h>
#include <vector>

int biggest_stadium( int N, std::vector<std::vector<int>> F ){
  int nempty = 0, l, c, border, prev, minl, maxl, minc, maxc, max_fl, max_fc;

  std::vector<int> fl( N, 0 ), fc( N, 0 );

  minl = N;
  maxl = -1;
  minc = N;
  maxc = -1;
  for( l = 0 ; l < N ; l++ )
    for( c = 0 ; c < N ; c++ )
      if( !F[l][c] ){
        nempty++;
        fl[l]++;
        fc[c]++;

        minl = l < minl ? l : minl;
        maxl = l > minl ? l : minl;
        minc = c < minc ? c : minc;
        maxc = c > minc ? c : minc;
      }

  for( l = 0 ; l < N ; l++ ){
    prev = 1;
    border = 0;
    for( c = 0 ; c < N ; c++ ){
      if( F[l][c] != prev ){
        prev = F[l][c];
        border++;
      }
    }

    if( border > 2 )
      return 1;
  }

  for( c = 0 ; c < N ; c++ ){
    prev = 1;
    border = 0;
    for( l = 0 ; l < N ; l++ ){
      if( F[l][c] != prev ){
        prev = F[l][c];
        border++;
      }
    }

    if( border > 2 )
      return 1;
  }

  max_fl = max_fc = 0;
  for( l = 0 ; l < N ; l++ ){
    max_fl = fl[l] > max_fl ? fl[l] : max_fl;
    max_fc = fc[l] > max_fc ? fc[l] : max_fc;
  }

  if( maxl - minl + 1 != max_fl )
    return 1;

  if( maxc - minc + 1 != max_fc )
    return 1;

  return nempty;

  // ignore the rest

  // since initial condition is met now we can focus on couting reachable pairs

  for( l = 0 ; l < N ; l++ )
    for( c = 0 ; c < N ; c++ )
      if( !F[l][c] ){
      }

  long long max_reachable = nempty;
  max_reachable *= (nempty - 1);
  max_reachable >>= 1; // (nempty choose 2)

  long long reachable_pairs = 0;

  for( l = 0 ; l < N ; l++ )
    for( c = 0 ; c < N ; c++ )
      if( !F[l][c] )
        reachable_pairs += ((long long)fl[l]) * fc[c];

  // all that is left is to count the pairs which allow 2 paths and remove them from the count

  reachable_pairs -= nempty;
  
  return reachable_pairs == max_reachable ? nempty : 1;
}

• 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