102: Ecological bin packing

This problem was suggested to me by uHunt. It was supposed to be again another easy problem. But upon first reading the problem, it is very intimidating. The problem starts off with an introduction about NP-problems. Big words that make the problem intimidating. I was thinking about dynamic programming algorithms or graph theory algorithms to solve the problem. I was indeed intimidated. In addition, the problem was quite lengthy.

Then I decided to search Google about the problem. Then upon first search, I saw the key words "Brute force". I thought about it a little and I thought that brute force may not be a bad idea after all. I counted the total number of brute force cases and there were only 6. So I decided to hard code them all since there were only 6 cases. For each case, compute for the number of movements. Then choose the case that has the minimum number of cases.

But there's a problem in choosing the case with the minimum number of movements. The problem constrains that the case chosen should be the one that comes out first in alphabetical sorting. Thus, I had to arrange the cases in such a way that the first case was the first in alphabetical order, and the last case was the last in alphabetical sorting. Here's my sample code:

#include

int main() {
  char line[100];
  int b0[3], b1[3], b2[3], x[6], i, j, min;
  
  while (fgets(line, 100, stdin)) {
    sscanf(line, "%d %d %d %d %d %d %d %d %d"
 , &b0[0], &b0[1], &b0[2]
 , &b1[0], &b1[1], &b1[2]
 , &b2[0], &b2[1], &b2[2]
    );
    
    // B = 0; G = 1; C = 2
    
    // 021 BCG
    x[0] = (b1[0] + b2[0]) + (b0[2] + b2[2]) + (b0[1] + b1[1]);

    // 012 BGC
    x[1] = (b1[0] + b2[0]) + (b0[1] + b2[1]) + (b0[2] + b1[2]);
    
    // 201 CBG
    x[2] = (b1[2] + b2[2]) + (b0[0] + b2[0]) + (b0[1] + b1[1]);

    // 210 CGB
    x[3] = (b1[2] + b2[2]) + (b0[1] + b2[1]) + (b0[0] + b1[0]);
    
    // 102 GBC
    x[4] = (b1[1] + b2[1]) + (b0[0] + b2[0]) + (b0[2] + b1[2]);

    // 120 GCB
    x[5] = (b1[1] + b2[1]) + (b0[2] + b2[2]) + (b0[0] + b1[0]);    
    
    for (i = 4, j = 5, min = x[5]; i >= 0; i -= 1) {
      if (x[i] <= min) {
      min = x[i];
      j = i;
      }
    }

//    for (i = 0; i < 6; i += 1) {
//      printf("%d: %d\n", i, x[i]);
//    }

    switch(j) {
      case 0: printf("BCG %d\n", x[j]);
              break;
      case 1: printf("BGC %d\n", x[j]);
              break;
      case 2: printf("CBG %d\n", x[j]);
              break;
      case 3: printf("CGB %d\n", x[j]);
              break;
      case 4: printf("GBC %d\n", x[j]);
              break;
      case 5: printf("GCB %d\n", x[j]);
              break;      
    }
  }
  
  return 0;
}