lc 312 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
DP Accepted
这道题目应该从逆向来思考，选择以某个气球为分割点，那么其左边部分和右边部分都要依赖与那个气球。dp[i][j]表示到最后只剩下i和j所产生金币的最大值，对于i和j之间的k，如果引爆k，那么dp[i][j]被更新成max(dp[i][j], nums[i]*nums[k]*nums[j]+dp[i][k]+dp[k][j])，所以想要得到这个数组最大金币值，需要在首尾各加上一个元素，值为1，此时答案即为dp[0][len-1]。
class Solution {
public:
int maxCoins(vector<int>& nums) {
if (nums.size() == 0) return 0;
nums.insert(nums.begin(), 1);
nums.insert(nums.end(), 1);
int len = nums.size();
vector<vector<int>> dp(len, vector<int>(len, 0));
for (int i = len-3; i >= 0; i--) {
for (int j = i+2; j < len; j++) {
for (int k = i+1; k < j; k++) {
dp[i][j] = max(dp[i][j], nums[i]*nums[k]*nums[j]+dp[i][k]+dp[k][j]);
}
}
}
return dp[0][len-1];
}
};