There are n piles of coins on a table. Each pile consists of a positive number of coins of assorted denominations.
In one move, you can choose any coin on top of any pile, remove it, and add it to your wallet.
Given a list piles, where piles[i] is a list of integers denoting the composition of the ith pile from top to bottom, and a positive integer k, return the maximum total value of coins you can have in your wallet if you choose exactly k coins optimally.
Example 1:
Input: piles = [[1,100,3],[7,8,9]], k = 2
Output: 101
Explanation:
The above diagram shows the different ways we can choose k coins.
The maximum total we can obtain is 101.
Example 2:
Input: piles = [[100],[100],[100],[100],[100],[100],[1,1,1,1,1,1,700]], k = 7
Output: 706
Explanation:
The maximum total can be obtained if we choose all coins from the last pile.
Constraints:
- n == piles.length
- 1 <= n <= 1000
- 1 <= piles[i][j] <= 105
- 1 <= k <= sum(piles[i].length) <= 2000
假设 dp[i][k]是当我们取到 piles[i]时还剩下 k 次机会所能拿到的最大值, dp[i][k] = max(
sum(piles[i][…1]) + dp[i+1][k-1],
sum(piles[i][…2]) + dp[i+1][k-2],
sum(piles[i][…3]) + dp[i+1][k-3],
…
sum(piles[i][…n]) + dp[i+1][k-n])
这里要注意一点是当前 pile 我们可以不取硬币直接跳过
use std::collections::HashMap;
impl Solution {
fn dp(piles: &Vec<Vec<i32>>, mut k: i32, i: usize, cache: &mut HashMap<(usize, i32), i32>) -> i32 {
if k == 0 || i == piles.len() {
return 0;
}
if let Some(&c) = cache.get(&(i, k)) {
return c;
}
let mut ans = 0;
let mut curr = 0;
ans = ans.max(Solution::dp(piles, k, i + 1, cache));
let mut curr_k = k;
for &v in &piles[i] {
curr += v;
curr_k -= 1;
if curr_k < 0 {
break;
}
let next = Solution::dp(piles, curr_k, i + 1, cache);
ans = ans.max(curr + next);
}
cache.insert((i, k), ans);
ans
}
pub fn max_value_of_coins(piles: Vec<Vec<i32>>, k: i32) -> i32 {
Solution::dp(&piles, k, 0, &mut HashMap::new())
}
}