2022-12-08：给定n棵树，和两个长度为n的数组a和b i号棵树的初始重量为a[i]，i号

2022-12-08：给定n棵树，和两个长度为n的数组a和b
i号棵树的初始重量为a[i]，i号树每天的增长重量为b[i]
你每天最多能砍1棵树，这天收益 = 砍的树初始重量 + 砍的树增长到这天的总增重
给定m，表示你有m天，返回m天内你获得的最大收益。

答案2022-12-08：

排序+贪心。

代码用rust编写。代码如下：

use std::iter::repeat;
fn main() {
    unsafe {
        let ins = [2, 2, 1, 10, 10, 1, 1, 2, 2, 8, 10, 2, 3];
        let mut ii = 0;
        let testCases = ins[ii];
        ii += 1;
        for i in 0..testCases {
            //
            let n = ins[ii];
            ii += 1;
            let m = ins[ii];
            ii += 1;
            for j in 0..n {
                tree[j as usize][0] = ins[ii];
                ii += 1;
            }
            for j in 0..n {
                tree[j as usize][1] = ins[ii];
                ii += 1;
            }
            let ans = max_weight(n, m);
            println!("ans = {}", ans);
        }
    }
}

static mut tree: [[i32; 2]; 250] = [[0; 2]; 250];
static mut dp: [[i32; 250]; 250] = [[0; 250]; 250];
// tree[][]
// i棵树，初始重量 ， tree[i][0]
// i棵树，每天的增长重量 ，tree[i][1]
fn max_weight(n: i32, m: i32) -> i32 {
    unsafe {
        //Arrays.sort(tree, 0, n, (o1, o2) -> o1[1] - o2[1]);
        tree[..n as usize].sort_by(|a, b| a[1].cmp(&b[1]));
        dp[0][0] = tree[0][0];
        for i in 1..n {
            dp[i as usize][0] = get_max(dp[(i - 1) as usize][0], tree[i as usize][0]);
        }
        for j in 1..m {
            dp[0][j as usize] = dp[0][(j - 1) as usize] + tree[0][1];
        }
        for i in 1..n {
            for j in 1..m {
                dp[i as usize][j as usize] = get_max(
                    dp[(i - 1) as usize][j as usize],
                    dp[(i - 1) as usize][(j - 1) as usize]
                        + tree[i as usize][0]
                        + tree[i as usize][1] * j,
                );
            }
        }
        return dp[(n - 1) as usize][(m - 1) as usize];
    }
}

fn get_max<T: Clone + Copy + std::cmp::PartialOrd>(a: T, b: T) -> T {
    if a > b {
        a
    } else {
        b
    }
}

执行结果如下：

左神java代码

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