2022-12-18：给定一个长度为n的二维数组graph，代表一张图，
graph[i] = {a,b,c,d} 表示i讨厌(a,b,c,d)，讨厌关系为双向的，
一共有n个人，编号0~n-1，
讨厌的人不能一起开会。
返回所有人能不能分成两组开会。
来自微软面试。
答案2022-12-18：
力扣785。并查集。
代码用rust编写。代码如下：
代码语言：javascript复制use rand::Rng;
use std::iter::repeat;
fn main() {
let mut graph: Vec<Vec<i32>> = vec![vec![1, 2, 3], vec![0, 2], vec![0, 1, 3], vec![0, 2]];
let ans = is_bipartite(&mut graph);
println!("ans = {}", ans);
}
fn is_bipartite(graph: &mut Vec<Vec<i32>>) -> bool {
let n = graph.len() as i32;
let mut uf = UnionFind::new(n);
for neighbours in graph.iter() {
for i in 1..neighbours.len() as i32 {
uf.union(neighbours[(i - 1) as usize], neighbours[i as usize]);
}
}
for i in 0..n {
for j in graph[i as usize].iter() {
if uf.same(i, *j) {
return false;
}
}
}
return true;
}
struct UnionFind {
f: Vec<i32>,
s: Vec<i32>,
h: Vec<i32>,
}
impl UnionFind {
pub fn new(n: i32) -> Self {
let mut f: Vec<i32> = repeat(0).take(n as usize).collect();
let mut s: Vec<i32> = repeat(0).take(n as usize).collect();
let mut h: Vec<i32> = repeat(0).take(n as usize).collect();
for i in 0..n {
f[i as usize] = i;
s[i as usize] = 1;
}
Self { f, s, h }
}
fn find(&mut self, mut i: i32) -> i32 {
let mut hi = 0;
while i != self.f[i as usize] {
self.h[hi as usize] = i;
hi = 1;
i = self.f[i as usize];
}
while hi > 0 {
hi -= 1;
self.f[self.h[hi as usize] as usize] = i;
}
return i;
}
pub fn same(&mut self, i: i32, j: i32) -> bool {
return self.find(i) == self.find(j);
}
pub fn union(&mut self, i: i32, j: i32) {
let mut fi = self.find(i);
let mut fj = self.find(j);
if fi != fj {
if self.s[fi as usize] >= self.s[fj as usize] {
self.f[fj as usize] = fi;
self.s[fi as usize] = self.s[fi as usize] self.s[fj as usize];
} else {
self.f[fi as usize] = fj;
self.s[fj as usize] = self.s[fi as usize] self.s[fj as usize];
}
}
}
}执行结果如下：
***
[左神java代码](https://github.com/algorithmzuo/weekly-problems/blob/main/src/class_2022_09_1_week/Code05_IsGraphBipartite.java)
