Burst Balloons

Given n balloons, indexed from 0 to n-1. Each balloon is painted with a number on it represented by array nums. You are asked to burst all the balloons. If the you burst balloon i you will get nums[left] * nums[i] * nums[right] coins. Here left and right are adjacent indices of i. After the burst, the left and right then becomes adjacent.

Find the maximum coins you can collect by bursting the balloons wisely.

Note:
(1) You may imagine nums[-1] = nums[n] = 1. They are not real therefore you can not burst them.
(2) 0 ≤ n ≤ 500, 0 ≤ nums[i] ≤ 100

Example:

Given [3, 1, 5, 8]

Return 167

    nums = [3,1,5,8] --> [3,5,8] -->   [3,8]   -->  [8]  --> []
   coins =  3*1*5      +  3*5*8    +  1*3*8      + 1*8*1   = 167


class Solution {
public:
    int maxCoins(vector<int>& nums) {
        int n = nums.size();
        int size = n+2;
        int num[size];
        for (int i = 0; i < n; i++)
        {
            num[i+1] = nums[i];
        }
        num[0] = 1;
        num[size-1] = 1;
        int buf[size][size];
        memset(buf, 0, sizeof(buf));
        for (int i = 2; i < size; i++)
        {
            for (int left = 0; left+i < size; left++)
            {
                int right = left+i;
                for (int j = left+1; j < right; j++)
                {
                     //该等式意味着在left和i之间，以及i与right之间的气球都已经被用完.
                     //所以这时获得的值是num[left]*num[j]*num[right].
                     //同时，由于在left和i之间的气球用完时，会用到num[j]的值，因此要加上buf[left][j].
                     //同理，也要加上buf[j][right].
                    buf[left][right] = max(buf[left][right], 
                                        buf[left][j] + num[left]*num[j]*num[right] + buf[j][right]);
                }
            }
        }

        return buf[0][size-1];
    }
};