Stone Game

Description

There is a stone game.At the beginning of the game the player picks n piles of stones in a line.

The goal is to merge the stones in one pile observing the following rules:

1. At each step of the game,the player can merge two adjacent piles to a new pile.
2. The score is the number of stones in the new pile.

You are to determine the minimum of the total score.

Example

Example 1:

Input: [3, 4, 3]
Output: 17

Example 2:

Input: [4, 1, 1, 4]
Output: 18
Explanation:
  1. Merge second and third piles => [4, 2, 4], score = 2
  2. Merge the first two piles => [6, 4]，score = 8
  3. Merge the last two piles => [10], score = 18

思路：

区间动态规划.

设定状态: f[i][j] 表示合并原序列 [i, j] 的石子的最小分数

状态转移: f[i][j] = min{f[i][k] + f[k+1][j]} + sum[i][j], sum[i][j] 表示原序列[i, j]区间的重量和

边界: f[i][i] = 0, f[i][i+1] = sum[i][i+1]

答案: f[0][n-1]

public class Solution {
    /**
     * @param A an integer array
     * @return an integer
     */
    int search(int l, int r, int[][] f, int[][] visit, int[][] sum) {
        if (visit[l][r] == 1)
            return f[l][r];
        if (l == r) {
            visit[l][r] = 1;
            return f[l][r];
        }

        f[l][r] = Integer.MAX_VALUE;
        for (int k = l; k < r; k++) {
            f[l][r] = Math.min(f[l][r], search(l, k, f, visit, sum) + search(k + 1, r, f, visit, sum) + sum[l][r]);
        }
        visit[l][r] = 1;
        return f[l][r];
    }

    public int stoneGame(int[] A) {
        if (A == null || A.length == 0) {
            return 0;
        }

        int n = A.length;

        // initialize
        int[][] f = new int[n][n];
        int[][] visit = new int[n][n];

        for (int i = 0; i < n; i++) {
            f[i][i] = 0;
        }

        // preparation
        int[][] sum = new int[n][n];
        for (int i = 0; i < n; i++) {
            sum[i][i] = A[i];
            for (int j = i + 1; j < n; j++) {
                sum[i][j] = sum[i][j - 1] + A[j];
            }
        }

        return search(0, n-1, f, visit, sum);
    }
}