2022-12-20:二狗买了一些小兵玩具,和大胖一起玩, 一共有n个小兵,这n个小兵拍成一列, 第i个小

2022-12-20:二狗买了一些小兵玩具,和大胖一起玩,
一共有n个小兵,这n个小兵拍成一列,
第i个小兵战斗力为hi,然后他们两个开始对小兵进行排列,
一共进行m次操作,二狗每次操作选择一个数k,将前k个小兵战斗力从小到大排列,
大胖每次操作选择一个数k,将前k个小兵战斗力从大到小排列,
问所有操作结束后,排列顺序什么样,
给定一个长度为n的数组arr,表示每个小兵的战斗力,
给定一个长度为m的数组op,
op[i] = { k , 0 }, 表示对前k个士兵执行从小到大的操作,
op[i] = { k , 1 }, 表示对前k个士兵执行从大到小的操作。
返回数组ans,表示最终的排列。
1 <= n, m <= 2 * 10^5,
-10 ^ 9<= arr[i] <= + 10^9。
来自百度。

答案2022-12-20:

单调栈+有序表。
rust语言里结构体的有序表需要实现Ord的trait。
时间复杂度O(M) + O(N*logN)。

rust代码如下:

use rand::Rng;
use std::cmp::Ordering;
use std::collections::BTreeSet;
use std::iter::repeat;
fn main() {
    let mut arr = vec![3, 2, 6, 7, 5, 1];
    let mut op = vec![vec![3, 0], vec![4, 1], vec![2, 0]];
    let ans2 = game2(&mut arr, &mut op);
    println!("ans2 = {:?}", ans2);
    let nn: i32 = 200;
    let mm: i32 = 100;
    let vv: i32 = 200;
    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 m: i32 = rand::thread_rng().gen_range(0, mm) + 1;
        let mut arr = random_array(n, vv);
        let mut op = random_op(m, n);
        let ans1 = game1(&mut arr, &mut op);
        let ans2 = game2(&mut arr, &mut op);
        if ans1 != ans2 {
            println!("出错了!");
            println!("i = {}", i);
            println!("ans1 = {:?}", ans1);
            println!("ans2 = {:?}", ans2);
            break;
        }
    }
    println!("测试结束");
}

// 暴力方法
// 为了验证
fn game1(arr: &mut Vec<i32>, op: &mut Vec<Vec<i32>>) -> Vec<i32> {
    let n = arr.len() as i32;
    let mut help: Vec<i32> = repeat(0).take(n as usize).collect();
    for i in 0..n {
        help[i as usize] = arr[i as usize];
    }
    for o in op.iter() {
        if o[1] == 0 {
            help[0..o[0] as usize].sort_by(|a, b| a.cmp(b));
        } else {
            help[0..o[0] as usize].sort_by(|a, b| b.cmp(a));
        }
    }
    let mut ans: Vec<i32> = repeat(0).take(n as usize).collect();
    for i in 0..n {
        ans[i as usize] = help[i as usize];
    }
    return ans;
}

// 正式方法
// 时间复杂度O(M) + O(N*logN)
fn game2(arr: &mut Vec<i32>, op: &mut Vec<Vec<i32>>) -> Vec<i32> {
    let n = arr.len() as i32;
    let m = op.len() as i32;
    let mut stack: Vec<i32> = repeat(0).take(m as usize).collect();
    let mut r = 0;
    for i in 0..m {
        while r != 0 && op[stack[(r - 1) as usize] as usize][0] <= op[i as usize][0] {
            r -= 1;
        }
        stack[r as usize] = i;
        r += 1;
    }
    let mut ans: Vec<i32> = repeat(0).take(n as usize).collect();
    let mut ansi = n - 1;
    let mut l = 0;
    while ansi >= op[stack[l as usize] as usize][0] {
        ans[ansi as usize] = arr[ansi as usize];
        ansi -= 1;
    }
    let mut sorted: BTreeSet<Number> = BTreeSet::new();
    for i in 0..op[stack[l as usize] as usize][0] {
        sorted.insert(Number::new(arr[i as usize], i));
    }
    while l != r {
        // 当前处理的指令
        let cur = &op[stack[l as usize] as usize];
        l += 1;
        if l != r {
            let mut next = &op[stack[l as usize] as usize];
            let num = cur[0] - next[0];
            if cur[1] == 0 {
                for i in 0..num {
                    ans[ansi as usize] = sorted.pop_last().unwrap().value;
                    ansi -= 1;
                }
            } else {
                for i in 0..num {
                    ans[ansi as usize] = sorted.pop_first().unwrap().value;
                    ansi -= 1;
                }
            }
        } else {
            if cur[1] == 0 {
                while sorted.len() > 0 {
                    ans[ansi as usize] = sorted.pop_last().unwrap().value;
                    ansi -= 1;
                }
            } else {
                while sorted.len() > 0 {
                    ans[ansi as usize] = sorted.pop_first().unwrap().value;
                    ansi -= 1;
                }
            }
        }
    }
    return ans;
}

struct Number {
    value: i32,
    index: i32,
}

impl Number {
    fn new(v: i32, i: i32) -> Self {
        Self { value: v, index: i }
    }
}

impl Ord for Number {
    fn cmp(&self, other: &Self) -> Ordering {
        if self.value != other.value {
            self.value.cmp(&other.value)
        } else {
            self.index.cmp(&other.index)
        }
    }
}

impl PartialOrd for Number {
    fn partial_cmp(&self, other: &Self) -> Option<Ordering> {
        Some(self.cmp(other))
    }
}

impl PartialEq for Number {
    fn eq(&self, other: &Self) -> bool {
        (self.value, &self.index) == (other.value, &other.index)
    }
}

impl Eq for Number {}

// 为了测试
fn random_array(n: i32, v: i32) -> Vec<i32> {
    let mut ans: Vec<i32> = repeat(0).take(n as usize).collect();
    for i in 0..n {
        ans[i as usize] = rand::thread_rng().gen_range(0, v) + 1;
    }
    return ans;
}

// 为了测试
fn random_op(m: i32, n: i32) -> Vec<Vec<i32>> {
    let mut ans: Vec<Vec<i32>> = repeat(repeat(0).take(2).collect())
        .take(m as usize)
        .collect();
    for i in 0..m {
        ans[i as usize][0] = rand::thread_rng().gen_range(0, n + 1);
        ans[i as usize][1] = rand::thread_rng().gen_range(0, 2);
    }
    return ans;
}

// 为了测试
fn is_equal(arr1: &mut Vec<i32>, arr2: &mut Vec<i32>) -> bool {
    if arr1.len() != arr2.len() {
        return false;
    }
    for i in 0..arr1.len() {
        if arr1[i as usize] != arr2[i as usize] {
            return false;
        }
    }
    return true;
}

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