312. Burst Balloons

最新推荐文章于 2025-03-22 19:50:36 发布

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最新推荐文章于 2025-03-22 19:50:36 发布

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本文链接：https://blog.youkuaiyun.com/qq_40127318/article/details/78812688

探讨了如何通过动态规划解决气球爆破问题，旨在最大化获取的硬币数。文章详细介绍了状态转移方程，并提供了C++实现代码。

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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

分析

令dp[i][j]表示[i,j]的所有气球都爆掉能能获得的最大硬币数量。

设k是[i,j]中最后一个爆炸的气球，i<=k<=j。

则dp[i][j]=max{dp[i][k-1]+dp[k+1][j]+nums[i-1]*dp[k]*nums[j+1]},i<=k<=j.

依次计算子区间长度为1、2、3...size()时的dp[i][j]

最大的硬币数量=dp[1][n]

代码

class Solution {
public:
    int maxCoins(vector<int>& nums) {
		int n=nums.size();
		nums.insert(nums.begin(),1);
		nums.push_back(1);
		int dp[n+2][n+2]={0};
		for (int len=1; len <= n; ++len)
			for (int left=1; left <= n-len+1; ++left) {
				int right=left+len-1;
				for (int k = left; k <= right; ++k)
					dp[left][right]=max(dp[left][right],dp[left][k-1]+dp[k+1][right]+nums[left-1]*nums[k]*nums[right+1]);
			}
		return dp[1][n];
	}
};