[leetcode] 312. Burst Balloons

與矩陣連乘問題很相似：

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:

• 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

最後一個扎破的氣球要乘以原區間外相鄰左右兩邊的值，沒有時則乘以1，將nums轉化爲 a[1000]，a[0] = a[n + 1] = 1，即可解決乘以1的問題，這題與矩陣連乘問題很相似，解決辦法也很相似。設 k 是最後一個扎破的氣球，dp[i][k - 1] 以及dp[k + 1][j]都是已經完全得到的具體數值，爲[]，那麼僅需加上 a[i - 1] * a[k] * a[j - 1]，i , j 分別爲所求區間的端點，L從1開始，經過判斷，對長度爲 1 的dp也可以按正確求得，注意 k = i; k <= j; 是正確的，雖然可能出現 k - 1 > i 的情況，但是由於已經初始化dp數組爲0，則不會產生影響。

class Solution {
public:
    int a[1000];
    int dp[501][501] = {0};
    int maxCoins(vector<int>& nums) {
        int n = nums.size();
        for(int i = 0; i < n; i++)
            a[i + 1] = nums[i];
        a[0] = 1;
        a[n + 1] = 1;
        for(int L = 1; L <= n; L++)
        {
            for(int i = 1; i <= n - L + 1; i++)
            {
                int j = i + L - 1;
                for(int k = i; k <= j; k++)
                {
                    dp[i][j] = max(dp[i][j], dp[i][k - 1] + a[i - 1] * a[k] * a[j + 1] + dp[k + 1][j]);
                }
            }
        }
        return dp[1][n];
    }
};