LintCode 593: Stone Game II (DP经典题!)

593. Stone Game II

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

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

At each step of the game,the player can merge two adjacent piles to a new pile.
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:
[1,1,4,4]
Output:18
Explanation:

1. Merge second and third piles => [2, 4, 4], score +2
2. Merge the first two piles => [6, 4]，score +6
3. Merge the last two piles => [10], score +10
   Example 2:

Input:
[1, 1, 1, 1]
Output:8
Explanation:

1. Merge first and second piles => [2, 1, 1], score +2
2. Merge the last two piles => [2, 2]，score +2
3. Merge the last two piles => [4], score +4

解法1：
DP。思路类似Stone Game。但要记得circular buffer的处理。dp[2n][2n], sums[2n]。
代码如下：

class Solution {
public:
    /**
     * @param A: An integer array
     * @return: An integer
     */
    int stoneGame2(vector &A) {
        int n = A.size();
        if (n == 0) return 0;
        
        vector> dp(n * 2, vector(n * 2, 0));
        vector sums(n * 2, 0);
        
        sums[0] = A[0];        
        for (int i = 1; i < n * 2; ++i) {
            sums[i] = sums[i - 1] + A[i % n];  
        }
        
        for (int len = 2; len <= n; ++len) {
            for (int i = 0; i + len - 1 < n * 2; ++i) {
                dp[i][i + len - 1] = INT_MAX;
                int sum = sums[i + len - 1] - sums[i - 1];
                for (int k = i; k < i + len - 1; ++k) {
                    dp[i][i + len - 1] = min(dp[i][i + len - 1], dp[i][k] + dp[k + 1][i + len - 1] + sum);
                }
            }
        }
        
        int result = INT_MAX;
        for (int i = 0; i < n; ++i) {
            result = min(result, dp[i][i + n - 1]);
        }
        
        return result;
        
    }
};