Submission #561478

#	Time	Username	Problem	Language	Result	Execution time	Memory
561478		Sweezy	Land of the Rainbow Gold (APIO17_rainbow)	C++17	0 / 100	2 ms	700 KiB

rainbow

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

using namespace std;

const int dx[4] = {1, -1, 0, 0};
const int dy[4] = {0, 0, 1, -1};

const int N = 100;
char table[N][N];

int n, m;

void init(int R, int C, int sr, int sc, int M, char *S) {
  n = R, m = C;
  table[sr][sc] = 1;
  for (int i = 0; i < M; i++) {
    if (S[i] == 'N') {
      sr--;
    } else if (S[i] == 'S') {
      sr++;
    } else if (S[i] == 'W') {
      sc--;
    } else {
      sc++;
    }
    table[sr][sc] = 1;
  }
}

bool is_valid(int x, int y, int lx, int rx, int ly, int ry) {
  return x >= lx && x <= rx && y >= ly && y <= ry;
}

bool value(int x, int y, int lx, int rx, int ly, int ry) {
  return x < lx || x > rx || y < ly || y > ry || table[x][y];
}

int colour(int ar, int ac, int br, int bc) {
  bool found = false;
  for (int i = ar; i <= br; i++) {
    for (int j = ac; j <= bc; j++) {
      if (table[i][j]) {
        ar = i;
        found = true;
        break;
      }
    }
    if (found) {
      break;
    }
  }
  if (!found) {
    return 1;
  }
  int e = 0, v = 0;
  int e_border = 0;
  for (int i = ar; i <= br; i++) {
    for (int j = ac; j <= bc; j++) {
      if (table[i][j]) {
        v++;
        for (int k = 0; k < 4; k++) {
          int ni = i + dx[k];
          int nj = j + dy[k];
          if (is_valid(ni, nj, ar, br, ac, bc) && table[ni][nj]) {
            e++;
          } else if (value(ni, nj, ar, br, ac, bc)) {
            e_border++;
          }
        }
      }
    }
  }
  e /= 2;
  v += 2 * (br - ar + 1 + bc - ac + 1) + 4;
  e += 2 * (br - ar + 1 + bc - ac + 1) + 4;
  e += e_border;

  // return e;

  int sub = 0;
  for (int i = ar - 1; i <= br; i++) {
    for (int j = ac - 1; j <= bc; j++) {
      sub += value(i, j, ar, br, ac, bc) & value(i + 1, j, ar, br, ac, bc) & value(i + 1, j + 1, ar, br, ac, bc) & value(i, j + 1, ar, br, ac, bc);
    }
  }

  return e - v + 1 - sub;
}

• Subtask 1: 0 / 11
• Subtask 2: 0 / 12
• Subtask 3: 0 / 24
• Subtask 4: 0 / 27
• Subtask 5: 0 / 26