2022-07-29：一共有n个人，从左到右排列，依次编号0~n-1， h[i]是第i个人的身高， v[i]是第i个人的分数， 要求从左到右选出一个子序列，在这个子序列中的人，从左到右身高是不下降的。

Posted 2022-08-07 福大大架构师每日一题

2022-07-29：一共有n个人，从左到右排列，依次编号0~n-1，
h[i]是第i个人的身高，
v[i]是第i个人的分数，
要求从左到右选出一个子序列，在这个子序列中的人，从左到右身高是不下降的。
返回所有符合要求的子序列中，分数最大累加和是多大。
n <= 10的5次方, 1 <= h[i] <= 10的9次方, 1 <= v[i] <= 10的9次方。
来自字节。

答案2022-07-29：

线段树。

代码用rust编写。代码如下：

use rand::Rng;
fn main() 
    let nn: i32 = 30;
    let vv: i32 = 100;
    let test_time: i32 = 20000;
    println!("测试开始");
    for _ in 0..test_time 
        let n = rand::thread_rng().gen_range(0, nn) + 1;
        let mut h = random_array(n, vv);
        let mut v = random_array(n, vv);
        if right(&mut h, &mut v) != max_sum(&mut h, &mut v) 
            println!("出错了!");
            break;
        
    
    println!("测试结束");


// 为了测试
// 绝对正确的暴力方法
fn right(h: &mut Vec<i32>, v: &mut Vec<i32>) -> i32 
    return process(h, v, 0, 0);


fn process(h: &mut Vec<i32>, v: &mut Vec<i32>, index: i32, pre_value: i32) -> i32 
    if index == h.len() as i32 
        return 0;
    
    let p1 = process(h, v, index + 1, pre_value);
    let p2 = if h[index as usize] >= pre_value 
        v[index as usize] + process(h, v, index + 1, h[index as usize])
     else 
        0
    ;
    return get_max(p1, p2);


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


// 正式方法
// 时间复杂度O(N * logN)
fn max_sum(h: &mut Vec<i32>, v: &mut Vec<i32>) -> i32 
    let n = h.len() as i32;
    let mut rank0 = h.clone();
    rank0.sort();
    let mut st = SegmentTree::new(n);
    for i in 0..n 
        let height = rank(&mut rank0, h[i as usize]);
        // 1~height max
        let t = st.max1(height);
        st.update1(height, t + v[i as usize]);
    
    return st.max1(n);


// [150, 152, 160, 175]  160
//   1    2    3    4
// 3
fn rank(rank0: &mut Vec<i32>, num: i32) -> i32 
    let mut l = 0;
    let mut r = rank0.len() as i32 - 1;
    let mut m = 0;
    let mut ans = 0;
    while l <= r 
        m = (l + r) / 2;
        if rank0[m as usize] >= num 
            ans = m;
            r = m - 1;
         else 
            l = m + 1;
        
    
    return ans + 1;


pub struct SegmentTree 
    pub n: i32,
    pub max: Vec<i32>,
    pub update: Vec<i32>,


impl SegmentTree 
    pub fn new(max_size: i32) -> Self 
        let n = max_size + 1;
        let mut max: Vec<i32> = vec![];
        let mut update: Vec<i32> = vec![];
        for _ in 0..n << 2 
            max.push(0);
            update.push(-1);
        
        Self  n, max, update 
    

    pub fn update1(&mut self, index: i32, c: i32) 
        self.update0(index, index, c, 1, self.n, 1);
    

    pub fn max1(&mut self, right: i32) -> i32 
        return self.max0(1, right, 1, self.n, 1);
    
    
    fn push_up(&mut self, rt: i32) 
        self.max[rt as usize] = get_max(
            self.max[(rt << 1) as usize],
            self.max[(rt << 1 | 1) as usize],
        );
    

    fn push_down(&mut self, rt: i32, _ln: i32, _rn: i32) 
        if self.update[rt as usize] != -1 
            self.update[(rt << 1) as usize] = self.update[rt as usize];
            self.max[(rt << 1) as usize] = self.update[rt as usize];
            self.update[(rt << 1 | 1) as usize] = self.update[rt as usize];
            self.max[(rt << 1 | 1) as usize] = self.update[rt as usize];
            self.update[rt as usize] = -1;
        
    

    fn update0(&mut self, ll: i32, rr: i32, cc: i32, l: i32, r: i32, rt: i32) 
        if ll <= l && r <= rr 
            self.max[rt as usize] = cc;
            self.update[rt as usize] = cc;
            return;
        
        let mid = (l + r) >> 1;
        self.push_down(rt, mid - l + 1, r - mid);
        if ll <= mid 
            self.update0(ll, rr, cc, l, mid, rt << 1);
        
        if rr > mid 
            self.update0(ll, rr, cc, mid + 1, r, rt << 1 | 1);
        
        self.push_up(rt);
    

    fn max0(&mut self, ll: i32, rr: i32, l: i32, r: i32, rt: i32) -> i32 
        if ll <= l && r <= rr 
            return self.max[rt as usize];
        
        let mid = (l + r) >> 1;
        self.push_down(rt, mid - l + 1, r - mid);
        let mut ans = 0;
        if ll <= mid 
            ans = get_max(ans, self.max0(ll, rr, l, mid, rt << 1));
        
        if rr > mid 
            ans = get_max(ans, self.max0(ll, rr, mid + 1, r, rt << 1 | 1));
        
        return ans;
    


// 为了测试
fn random_array(n: i32, v: i32) -> Vec<i32> 
    let mut ans: Vec<i32> = vec![];
    for _ in 0..n 
        ans.push(rand::thread_rng().gen_range(0, v) + 1);
    
    return ans;

执行结果如下：

---

左神java代码