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

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

Note:

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

Example:

Input: [3,1,5,8]
Output: 167 
Explanation: 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
 

解法
對第i個位置消去，遞迴之後的結果是arr[l-1]*arr[i]*arr[r+1]
dp[l][r] = max(dp[l][r], dp[l][i-1]+dp[i+1][j]+arr[l-1]*arr[r+1]*arr[i]

class Solution {
    int dp[505][505];
    int n;
    vector<int> arr;
    int dfs(int l, int r) {
        if(l>r) return 0;
        if(dp[l][r] > 0)
            return dp[
 
l][r];
        for(int i=l;i<=r;i++) {
            int left = l-1<0?1:arr[l-1];
            int right = r+1>=n?1:arr[r+1];
            dp[l][r] = max(dp[l][r], dfs(l, i-1) + dfs(i+1, r) + arr[i]*left*right);
        }
        return dp[l][r];
    }
public:
    int maxCoins(vector<int>& nums) {
        arr = nums;
        n = nums.size();
        return dfs(0, n-1);
    }
};