2023-02-16：两种颜色的球，蓝色和红色，都按1～n编号，共计2n个，
为方便放在一个数组中，红球编号取负，篮球不变，并打乱顺序，
要求同一种颜色的球按编号升序排列，可以进行如下操作：
交换相邻两个球，求最少操作次数。
[3,-3,1,-4,2,-2,-1,4]、
最终交换结果为：
[1,2,3,-1,-2,-3,-4,4]。
最少交换次数为10，
n <= 1000。
答案2023-02-16：
动态规划，IndexTree。
代码用rust编写。代码如下：
use std::collections::HashMap;
use std::iter::repeat;
fn main() {
let mut arr = vec![3, -3, 1, -4, 2, -2, -1, 4];
println!("{}", min_swaps(&mut arr));
}
// [3,-3,1,-4,2,-2,-1,4]
// -3 -4 -2 -1 -> -1 -2 -3 -4
// 3 1 2 4 -> 1 2 3 4
// 这个题写对数器太麻烦了
// 所以这就是正式解
fn min_swaps(arr: &mut Vec<i32>) -> i32 {
let n = arr.len() as i32;
let mut map: HashMap<i32, i32> = HashMap::new();
let mut top_a = 0;
let mut top_b = 0;
for i in 0..n {
if arr[i as usize] > 0 {
top_a = get_max(top_a, arr[i as usize]);
} else {
top_b = get_max(top_b, abs(arr[i as usize]));
}
map.insert(arr[i as usize], i);
}
let mut it = IndexTree::new(n);
for i in 0..n {
it.add(i, 1);
}
return f(top_a, top_b, &mut it, n - 1, &mut map);
}
// 可以改二维动态规划！
// 因为it的状态，只由topA和topB决定
// 所以it的状态不用作为可变参数！
fn f(top_a: i32, top_b: i32, it: &mut IndexTree, end: i32, map: &mut HashMap<i32, i32>) -> i32 {
if top_a == 0 && top_b == 0 {
return 0;
}
let mut p1 = i32::MAX;
let mut p2 = i32::MAX;
let mut index: i32;
let mut cost: i32;
let mut next: i32;
if top_a != 0 {
index = *map.get(&top_a).unwrap();
cost = it.sum(index, end) - 1;
it.add(index, -1);
next = f(top_a - 1, top_b, it, end, map);
it.add(index, 1);
p1 = cost next;
}
if top_b != 0 {
index = *map.get(&(-top_b)).unwrap();
cost = it.sum(index, end) - 1;
it.add(index, -1);
next = f(top_a, top_b - 1, it, end, map);
it.add(index, 1);
p2 = cost next;
}
return get_min(p1, p2);
}
fn get_max<T: Clone Copy std::cmp::PartialOrd>(a: T, b: T) -> T {
if a > b {
a
} else {
b
}
}
fn get_min<T: Clone Copy std::cmp::PartialOrd>(a: T, b: T) -> T {
if a < b {
a
} else {
b
}
}
fn abs(a: i32) -> i32 {
if a < 0 {
-a
} else {
a
}
}
struct IndexTree {
tree: Vec<i32>,
nn: i32,
}
impl IndexTree {
pub fn new(size: i32) -> Self {
Self {
tree: repeat(0).take((size 1) as usize).collect(),
nn: size,
}
}
pub fn add(&mut self, mut i: i32, v: i32) {
i = 1;
while i <= self.nn {
self.tree[i as usize] = v;
i = i & -i;
}
}
pub fn sum(&mut self, l: i32, r: i32) -> i32 {
return if l == 0 {
self.sum0(r 1)
} else {
self.sum0(r 1) - self.sum0(l)
};
}
fn sum0(&mut self, mut index: i32) -> i32 {
let mut ans = 0;
while index > 0 {
ans = self.tree[index as usize];
index -= index & -index;
}
return ans;
}
}
