编程练习（第十二周）

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这是一个LeetCode上的困难级别问题，编号为312，名为"Burst Balloons"。玩家需要爆掉由数字标记的气球，并根据爆掉的顺序获取硬币奖励。目标是找到最大化的硬币收集策略。给定一个包含0到500之间数字的数组，每爆掉一个气球会得到相邻数字的乘积作为奖励。问题要求使用智慧选择爆破顺序以最大化收益。示例给出了一组输入和期望的输出。解题思路通常涉及动态规划的方法。

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题目来源于:https://leetcode.com

312. Burst Balloons

Description Hints Submissions Solutions

Total Accepted: 24690
Total Submissions: 58469
Difficulty: Hard
Contributor: LeetCode

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

题解：

class Solution {
public:
    vector<vector<int>> d;
    int maxCoins(vector<int>& nums) {
        if(nums.empty()){
            return 0;
        }
        int size = nums.size();
        nums.insert(nums.begin(), 1);
        nums.insert(nums.end(), 1);
        d = vector<vector<int>>(size + 2, vector<int>(size + 2, 0));
        for(int len = 1; len <= size; ++len){
            for(int start = 1;start + len - 1 <= size; ++start){
                int end = start + len - 1;
                for(int k = start;k <= end; ++k){// k is the last one to be popped
                    d[start][end] = max(d[start][end], d[start][k - 1] + d[k + 1][end] + nums[start - 1] * nums[k] * nums[end + 1]);
                }
            }
        }
        return d[1][size];
    }
};