2022-08-02：小红拿到了一个大立方体，该大立方体由111的小方块拼成，初始每个小方块都是白色。
小红可以每次选择一个小方块染成红色，
每次小红可能选择同一个小方块重复染色，
每次染色以后，你需要帮小红回答出当前的白色连通块数，
如果两个小方块共用同一个面，且颜色相同，则它们是连通的，
给定n、m、h，表示大立方体的长、宽、高，
给定k次操作，每一次操作用(a, b, c)表示在大立方体的该位置进行染色。
返回长度为k的数组，表示每一次操作后，白色方块的连通块数。
来自网易。
答案2022-08-02：
并查集。时光倒流。
代码用rust编写。代码如下：
代码语言：rust复制use rand::Rng;
fn main() {
let size: i32 = 10;
let test_time: i32 = 5000;
println!("测试开始");
for _ in 0..test_time {
let n = rand::thread_rng().gen_range(0, size) 1;
let m = rand::thread_rng().gen_range(0, size) 1;
let h = rand::thread_rng().gen_range(0, size) 1;
let mut ops = random_ops(n, m, h);
let ans1 = blocks1(n, m, h, &mut ops);
let ans2 = blocks2(n, m, h, &mut ops);
if ans1 != ans2 {
println!("ans1 = {:?}", ans1);
println!("ans2 = {:?}", ans2);
println!("出错了!");
break;
}
}
println!("测试结束");
}
// 暴力方法
// 时间复杂度(k * n * m * h);
fn blocks1(n: i32, m: i32, h: i32, ops: &mut Vec<Vec<i32>>) -> Vec<i32> {
let mut k = ops.len() as i32;
//int[][][] cube = new int[n][m][h];
let mut cube: Vec<Vec<Vec<i32>>> = vec![];
for i in 0..n {
cube.push(vec![]);
for j in 0..m {
cube[i as usize].push(vec![]);
for _ in 0..h {
cube[i as usize][j as usize].push(0);
}
}
}
let mut value = 1;
let mut ans: Vec<i32> = vec![];
for _ in 0..k {
ans.push(0);
}
for i in 0..k {
cube[ops[i as usize][0] as usize][ops[i as usize][1] as usize]
[ops[i as usize][2] as usize] = -1;
for x in 0..n {
for y in 0..m {
for z in 0..h {
if cube[x as usize][y as usize][z as usize] != -1
&& cube[x as usize][y as usize][z as usize] != value
{
ans[i as usize] = 1;
infect(&mut cube, x, y, z, value);
}
}
}
}
value = 1;
}
return ans;
}
fn infect(cube: &mut Vec<Vec<Vec<i32>>>, a: i32, b: i32, c: i32, change: i32) {
if a < 0
|| a == cube.len() as i32
|| b < 0
|| b == cube[0].len() as i32
|| c < 0
|| c == cube[0][0].len() as i32
|| cube[a as usize][b as usize][c as usize] == -1
|| cube[a as usize][b as usize][c as usize] == change
{
return;
}
cube[a as usize][b as usize][c as usize] = change;
infect(cube, a - 1, b, c, change);
infect(cube, a 1, b, c, change);
infect(cube, a, b - 1, c, change);
infect(cube, a, b 1, c, change);
infect(cube, a, b, c - 1, change);
infect(cube, a, b, c 1, change);
}
// 最优解
// O(k n * m * h)
fn blocks2(n: i32, m: i32, h: i32, ops: &mut Vec<Vec<i32>>) -> Vec<i32> {
let mut k = ops.len() as i32;
//int[][][] red = new int[n][m][h];
let mut red: Vec<Vec<Vec<i32>>> = vec![];
for i in 0..n {
red.push(vec![]);
for j in 0..m {
red[i as usize].push(vec![]);
for _ in 0..h {
red[i as usize][j as usize].push(0);
}
}
}
for op in ops.iter() {
red[op[0] as usize][op[1] as usize][op[2] as usize] = 1;
}
let mut uf = UnionFind::new(n, m, h, &mut red);
let mut ans: Vec<i32> = vec![];
for _ in 0..k {
ans.push(0);
}
let mut i = k - 1;
while i >= 0 {
ans[i as usize] = uf.sets;
let mut x = ops[i as usize][0];
let mut y = ops[i as usize][1];
let mut z = ops[i as usize][2];
red[x as usize][y as usize][z as usize] -= 1;
if red[x as usize][y as usize][z as usize] == 0 {
// x, y ,z 这个格子，变白，建立自己的小集合
// 然后6个方向，集合该合并合并
uf.finger(x, y, z);
}
i -= 1;
}
return ans;
}
pub struct UnionFind {
n: i32,
m: i32,
h: i32,
father: Vec<i32>,
size: Vec<i32>,
help: Vec<i32>,
sets: i32,
}
impl UnionFind {
fn new(a: i32, b: i32, c: i32, red: &mut Vec<Vec<Vec<i32>>>) -> UnionFind {
let mut n = a;
let mut m = b;
let mut h = c;
let mut len = n * m * h;
let mut father: Vec<i32> = vec![];
let mut size: Vec<i32> = vec![];
let mut help: Vec<i32> = vec![];
for _ in 0..len {
father.push(0);
size.push(0);
help.push(0);
}
let mut ans = UnionFind {
n,
m,
h,
father,
size,
help,
sets: 0,
};
for x in 0..n {
for y in 0..m {
for z in 0..h {
if red[x as usize][y as usize][z as usize] == 0 {
ans.finger(x, y, z);
}
}
}
}
return ans;
}
pub fn finger(&mut self, x: i32, y: i32, z: i32) {
// x，y，z
// 一维数值
let mut i = self.index(x, y, z);
self.father[i as usize] = i;
self.size[i as usize] = 1;
self.sets = 1;
self.union(i, x - 1, y, z);
self.union(i, x 1, y, z);
self.union(i, x, y - 1, z);
self.union(i, x, y 1, z);
self.union(i, x, y, z - 1);
self.union(i, x, y, z 1);
}
fn index(&mut self, x: i32, y: i32, z: i32) -> i32 {
return z * self.n * self.m y * self.n x;
}
fn union(&mut self, mut i: i32, x: i32, y: i32, z: i32) {
if x < 0 || x == self.n || y < 0 || y == self.m || z < 0 || z == self.h {
return;
}
let mut j = self.index(x, y, z);
if self.size[j as usize] == 0 {
return;
}
i = self.find(i);
j = self.find(j);
if i != j {
if self.size[i as usize] >= self.size[j as usize] {
self.father[j as usize] = i;
self.size[i as usize] = self.size[j as usize];
} else {
self.father[i as usize] = j;
self.size[j as usize] = self.size[i as usize];
}
self.sets -= 1;
}
}
fn find(&mut self, mut i: i32) -> i32 {
let mut s = 0;
while i != self.father[i as usize] {
self.help[s] = i;
s = 1;
i = self.father[i as usize];
}
while s > 0 {
s -= 1;
self.father[self.help[s] as usize] = i;
}
return i;
}
}
// 为了测试
fn random_ops(n: i32, m: i32, h: i32) -> Vec<Vec<i32>> {
let mut size = rand::thread_rng().gen_range(0, n * m * h) 1;
let mut ans: Vec<Vec<i32>> = vec![];
for i in 0..size {
ans.push(vec![]);
for _ in 0..3 {
ans[i as usize].push(0);
}
}
for i in 0..size {
ans[i as usize][0] = rand::thread_rng().gen_range(0, n);
ans[i as usize][1] = rand::thread_rng().gen_range(0, m);
ans[i as usize][2] = rand::thread_rng().gen_range(0, h);
}
return ans;
}执行结果如下：
在这里插入图片描述左神java代码
