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

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:

在这里插入图片描述


左神java代码

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