2022-12-30：某天小美进入了一个迷宫探险，根据地图所示，这个迷宫里有无数个房间 序号分别为123...入口房间的序号为1 任意序号为正整数x的房间，都与序号 2*x 和 2*x + 1

Posted 2023-01-02 福大大架构师每日一题

2022-12-30：某天小美进入了一个迷宫探险，根据地图所示，这个迷宫里有无数个房间
序号分别为1、2、3、…入口房间的序号为1
任意序号为正整数x的房间，都与序号 2x 和 2x + 1 的房间之间各有一条路径
但是这些路径是单向的，即只能从序号为x的房间去到序号为 2x 或 2x+1 的房间
而不能从 2x 或 2x+1 的房间去到序号为x的房间
在任何时刻小美都可以选择结束探险并离开迷宫，但是离开之后将无法再次进入迷宫
小美还提前了解了迷宫中宝藏的信息
已知宝藏共有n个，其中第i个宝藏在序号为pi的房间，价值为wi
且一个房间中可能有多个宝藏
小美为了得到更多的宝藏，需要精心规划路线，她找到你帮忙
想请你帮她计算一下，能获得的宝藏价值和最大值为多少
第一行一个正整数n，表示宝藏数量。
第二行为n个正整数p1, p2,… pn，其中pi表示第 i 个宝藏在序号为pi的房间。
第三行为n个正整数w1, w2,… wn，其中wi表示第i个宝藏的价值为wi。
1 <= n <= 40000, 1 <= pi < 2^30, 1 <= wi <= 10^6。
来自美团。

答案2022-12-30：

动态规划，利用图来优化枚举。
时间复杂度：O(N)。
空间复杂度：O(N)。

代码用rust编写。代码如下：

use rand::Rng;
use std::collections::HashMap;
use std::iter::repeat;
fn main() 
    let nn: i32 = 100;
    let pp: i32 = 5000;
    let ww: i32 = 5000;
    let test_time: i32 = 5000;
    println!("测试开始");
    for i in 0..test_time 
        let n: i32 = rand::thread_rng().gen_range(0, nn) + 1;
        let mut p = random_array(n, pp);
        let mut w = random_array(n, ww);
        let ans1 = max_money1(n, &mut p, &mut w);
        let ans2 = max_money2(n, &mut p, &mut w);
        if ans1 != ans2 
            println!("出错了!");
            println!("i = ", i);
            println!("ans1 = :?", ans1);
            println!("ans2 = :?", ans2);
            break;
        
    
    println!("测试结束");


// 为了测试
// 普通动态规划
fn max_money1(n: i32, p: &mut Vec<i32>, w: &mut Vec<i32>) -> i32 
    let mut rooms: Vec<Vec<i32>> = repeat(repeat(0).take(2).collect())
        .take(n as usize)
        .collect();
    for i in 0..n 
        rooms[i as usize][0] = p[i as usize];
        rooms[i as usize][1] = w[i as usize];
    
    rooms.sort_by(|a, b| a[0].cmp(&b[0]));
    let mut ans = 0;
    let mut dp: Vec<i32> = repeat(-1).take(n as usize).collect();
    for i in 0..n 
        ans = get_max(ans, process1(i, &mut rooms, n, &mut dp));
    
    return ans;


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


fn process1(index: i32, rooms: &mut Vec<Vec<i32>>, n: i32, dp: &mut Vec<i32>) -> i32 
    if dp[index as usize] != -1 
        return dp[index as usize];
    
    let mut next = 0;
    for i in index + 1..n 
        if reach(rooms[index as usize][0], rooms[i as usize][0]) 
            next = get_max(next, process1(i, rooms, n, dp));
        
    
    let mut ans = rooms[index as usize][1] + next;
    dp[index as usize] = ans;
    return dp[index as usize];


fn reach(from: i32, mut to: i32) -> bool 
    while to >= from 
        if from == to 
            return true;
         else 
            to /= 2;
        
    
    return false;


// 正式方法
// 时间复杂度O(N)的动态规划
// 利用图来优化枚举
fn max_money2(n: i32, p: &mut Vec<i32>, w: &mut Vec<i32>) -> i32 
    let mut rooms: Vec<Vec<i32>> = repeat(repeat(0).take(2).collect())
        .take(n as usize)
        .collect();
    for i in 0..n 
        rooms[i as usize][0] = p[i as usize];
        rooms[i as usize][1] = w[i as usize];
    
    rooms.sort_by(|a, b| a[0].cmp(&b[0]));
    let mut first: HashMap<i32, i32> = HashMap::new();
    let mut graph: Vec<Vec<i32>> = vec![];
    for i in 0..n 
        let mut to = rooms[i as usize][0];
        while to > 0 
            if first.contains_key(&to) 
                graph[*first.get(&to).unwrap() as usize].push(i);
                break;
             else 
                to >>= 1;
            
        
        graph.push(vec![]);
        if !first.contains_key(&rooms[i as usize][0]) 
            first.insert(rooms[i as usize][0], i);
        
    
    let mut ans = 0;
    let mut dp: Vec<i32> = repeat(0).take(n as usize).collect();
    let mut i = n - 1;
    while i >= 0 
        let mut post = 0;
        for next in graph[i as usize].iter() 
            if rooms[*next as usize][0] == rooms[i as usize][0] 
                dp[i as usize] += dp[*next as usize];
             else 
                post = get_max(post, dp[*next as usize]);
            
        
        dp[i as usize] += post + rooms[i as usize][1];
        ans = get_max(ans, dp[i as usize]);
        i -= 1;
    
    return ans;


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