6 立方体成像

有一个 n * n * n 的立方体，其中一些单位立方体已经缺失（剩下部分不一定连通）。每个单位立方体重 1 克，且被涂上单一的颜色（即 6 个面的一颜色相同）。给出前、左、后、右、顶、底 6 个视图，你的任务是判断这个屋里剩下的最大的重量。

 

// LA2995 Image is Everything
// Rujia Liu
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

#define REP(i,n) for(int i = 0; i < (n); i++)

const int maxn = 10;
int n;
char pos[maxn][maxn][maxn];
char view[6][maxn][maxn];

char read_char() {
  char ch;
  for(;;) {
    ch = getchar();
    if((ch >= 'A' && ch <= 'Z') || ch == '.') return ch;
  }
}

void get(int k, int i, int j, int len, int &x, int &y, int &z)
{
  if (k == 0) { x = len; y = j; z = i; } 
  if (k == 1) { x = n - 1 - j; y = len; z = i; }
  if (k == 2) { x = n - 1 - len; y = n - 1 - j; z = i; }
  if (k == 3) { x = j; y = n - 1 - len; z = i; }
  if (k == 4) { x = n - 1 - i; y = j; z = len; }
  if (k == 5) { x = i; y = j; z = n - 1 - len; }
}

int main() {
  while(scanf("%d", &n) == 1 && n) {
    REP(i,n) REP(k,6) REP(j,n) view[k][i][j] = read_char();
    REP(i,n) REP(j,n) REP(k,n) pos[i][j][k] = '#';

    REP(k,6) REP(i,n) REP(j,n) if (view[k][i][j] == '.')
      REP(p,n) {
        int x, y, z;
        get(k, i, j, p, x, y, z);
        pos[x][y][z] = '.';
      }

    for(;;) {
      bool done = true;
      REP(k,6) REP(i,n) REP(j,n) if (view[k][i][j] != '.') {
        REP(p,n) {
          int x, y, z;
          get(k, i, j, p, x, y, z);
          if (pos[x][y][z] == '.') continue;
          if (pos[x][y][z] == '#') {
            pos[x][y][z] = view[k][i][j];
            break;
          }
          if (pos[x][y][z] == view[k][i][j]) break;
          pos[x][y][z] = '.';
          done = false;
        }
      }
      if(done) break;
    }

    int ans = 0;
    REP(i,n) REP(j,n) REP(k,n)
      if (pos[i][j][k] != '.') ans ++;

    printf("Maximum weight: %d gram(s)\n", ans);
  }
  return 0;
}