TopCoder Problem Statement - MutaliskEasy |

Single Round Match 658 Round 1 - Division II, Level Two |

Given attacks that do 9,3, and 1 point of damage, determine the fewest number of rounds needed to take out an opponent.

1 public class MutaliskEasy { 2 3 private static final int[] ATTACKS = {9, 3, 1}; 4 private static final int MAX_HP = 60; 5 6 int[][][] results = new int[MAX_HP + 1][MAX_HP + 1][MAX_HP + 1]; 7 8 public int minimalAttacks(int[] x) { 9 10 // Nothing to do. 11 if (x.length == 0) return 0; 12 13 // Just divide by the maximum attack value. 14 if (x.length == 1) { 15 if (x[0] % ATTACKS[0] == 0) { 16 return x[0] / ATTACKS[0]; 17 } else { 18 return x[0] / ATTACKS[0] + 1; 19 } 20 } 21 22 /* 23 * The problem does not fall into one of the easy cases, 24 * so we'll have to do some work... 25 */ 26 27 // Initialize results. 28 for (int i = 0; i <= MAX_HP; i++) { 29 for (int j = 0; j <= MAX_HP; j++) { 30 for (int k = 0; k <= MAX_HP; k++) { 31 results[i][j][k] = -1; 32 } 33 } 34 } 35 results[0][0][0] = 0; 36 37 /* 38 * We'll treat a length of 2 the same as if it had 3. Just 39 * give the 3rd element a value of 0. 40 */ 41 if (x.length == 2) return minimalAttacks(x[0], x[1], 0); 42 if (x.length == 3) return minimalAttacks(x[0], x[1], x[2]); 43 44 return -1; // Length of x must be >= 0 and <= 3 45 46 } 47 48 private int minimalAttacks(int x, int y, int z) { 49 50 // Don't allow values to go below 0. 51 if (x < 0) x = 0; 52 if (y < 0) y = 0; 53 if (z < 0) z = 0; 54 55 56 // If we've already calculated this result, just return it. 57 if (results[x][y][z] >= 0) return results[x][y][z]; 58 59 int minAttacks = Integer.MAX_VALUE; 60 61 // Try all possible combinations. 62 for (int i = 0; i < ATTACKS.length; i++) { 63 64 for (int j = 0; j < ATTACKS.length; j++) { 65 if (j == i) continue; 66 67 for (int k = 0; k < ATTACKS.length; k++) { 68 if ((k == i) || (k == j)) continue; 69 70 minAttacks = Math.min(minAttacks, 71 minimalAttacks( 72 x - ATTACKS[i], 73 y - ATTACKS[j], 74 z - ATTACKS[k])); 75 } 76 } 77 } 78 79 results[x][y][z] = minAttacks + 1; 80 return minAttacks + 1; 81 82 } 83 84 }

My first approach was a simple greedy algorithm. I sorted the opponents according to the amount of hit points they had remaining, and then assigned the biggest attack to the opponent with most hit points remaining, the next biggest attack to the next strongest opponent, and the last attack to the third strongest opponent. Then I re-sorted the opponents, and repeated. As the first test shows, this approach does not work.

Instead, we'll need to try every possible combination.

The minimalAttacks(int,int,int) method begins by ensuring we don't allow the hit point values to go below 0. Then it checks to see if we've already solved for this combination of x, y, and z. If so, then we can just return that value This technique is called memoization, and allows up to skip the calculations that follow if they've already been carried out.

If we don't already have a result, then the loops on lines 62-77 will recursively call minimalAttacks with every possible combination of assigning attacks to the various opponents. We keep track of the best result in the minAttacks variable, and then store this in the results matrix before returning it.

The minimalAttacks(int[]) method checks for a few easy cases (length of x is 0 or 1), and failing that initializes the results matrix, and then uses minimalAttacks(int, int, int) to calculate the result.

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