2022-11-04：給定一個(gè)正數(shù)n，表示有多少個(gè)節(jié)點(diǎn)
給定一個(gè)二維數(shù)組edges，表示所有無(wú)向邊
edges[i] = {a, b} 表示a到b有一條無(wú)向邊
edges一定表示得是一個(gè)無(wú)環(huán)無(wú)向圖，也就是樹結(jié)構(gòu)
每個(gè)節(jié)點(diǎn)可以染1、2、3三種顏色。
要求 : 非葉節(jié)點(diǎn)得相鄰點(diǎn)一定要至少有兩種和自己不同顏色得點(diǎn)。
返回一種達(dá)標(biāo)得染色方案，也就是一個(gè)數(shù)組，表示每個(gè)節(jié)點(diǎn)得染色狀況。
1 <= 節(jié)點(diǎn)數(shù)量 <= 10得5次方。
來(lái)自米哈游。
答案2022-11-04：
生成圖，選一個(gè)頭節(jié)點(diǎn)，深度優(yōu)先染色。
代碼用rust編寫。代碼如下：
use std::{iter::repeat, vec};use rand::Rng;fn main() { let nn: i32 = 100; let test_time: i32 = 1000; println!("測(cè)試開始"); for i in 0..test_time { let n = rand::thread_rng().gen_range(0, nn) + 1; let mut edges = random_edges(n); let mut ans = dye(n, &mut edges); if !right_answer(n, &mut edges, &mut ans) { println!("出錯(cuò)了!{}", i); break; } } println!("測(cè)試結(jié)束");}// 1 2 3 1 2 3 1 2 3const RULE1: [i32; 3] = [1, 2, 3];// // 1 3 2 1 3 2 1 3 2const RULE2: [i32; 3] = [1, 3, 2];fn dye(n: i32, edges: &mut Vec<Vec<i32>>) -> Vec<i32> { let mut graph: Vec<Vec<i32>> = vec![]; // 0 : { 2, 1 } // 1 : { 0 } // 2 : { 0 } for _i in 0..n { graph.push(vec![]); } for edge in edges.iter() { // 0 -> 2 // 1 -> 0 graph[edge[0] as usize].push(edge[1]); graph[edge[1] as usize].push(edge[0]); } // 選一個(gè)頭節(jié)點(diǎn)！ let mut head = -1; for i in 0..n { if graph[i as usize].len() >= 2 { head = i; break; } } // graph // head let mut colors: Vec<i32> = repeat(0).take(n as usize).collect(); if head == -1 { // 兩個(gè)點(diǎn)，互相連一下 // 把colors，所有位置，都設(shè)置成1 colors = repeat(1).take(n as usize).collect(); } else { // dfs 染色了！ colors[head as usize] = 1; let h = graph[head as usize][0]; dfs(&mut graph, h, 1, &RULE1, &mut colors); for i in 1..graph[head as usize].len() as i32 { let h = graph[head as usize][i as usize]; dfs(&mut graph, h, 1, &RULE2, &mut colors); } } return colors;}// 整個(gè)圖結(jié)構(gòu)，都在graph// 當(dāng)前來(lái)到得節(jié)點(diǎn)，是head號(hào)節(jié)點(diǎn)// head號(hào)節(jié)點(diǎn)，在level層// 染色得規(guī)則，rule {1,2,3...} {1,3,2...}// 做得事情：以head為頭得整顆樹，每個(gè)節(jié)點(diǎn)，都染上顏色// 填入到colors數(shù)組里去fn dfs(graph: &mut Vec<Vec<i32>>, head: i32, level: i32, rule: &[i32; 3], colors: &mut Vec<i32>) { colors[head as usize] = rule[(level % 3) as usize]; for next in graph[head as usize].clone().iter() { if colors[*next as usize] == 0 { dfs(graph, *next, level + 1, rule, colors); } }}// 為了測(cè)試// 生成無(wú)環(huán)無(wú)向圖fn random_edges(n: i32) -> Vec<Vec<i32>> { let mut order: Vec<i32> = repeat(0).take(n as usize).collect(); for i in 0..n { order[i as usize] = i; } let mut i = n - 1; while i >= 0 { order.swap(i as usize, rand::thread_rng().gen_range(0, i + 1) as usize); i -= 1; } let mut edges: Vec<Vec<i32>> = repeat(repeat(0).take(2).collect()) .take((n - 1) as usize) .collect(); for i in 1..n { edges[(i - 1) as usize][0] = order[i as usize]; edges[(i - 1) as usize][1] = order[rand::thread_rng().gen_range(0, i) as usize]; } return edges;}// 為了測(cè)試fn right_answer(n: i32, edges: &mut Vec<Vec<i32>>, colors: &mut Vec<i32>) -> bool { let mut graph: Vec<Vec<i32>> = vec![]; for _i in 0..n { graph.push(vec![]); } for edge in edges.iter() { graph[edge[0] as usize].push(edge[1]); graph[edge[1] as usize].push(edge[0]); } let mut has_colors: Vec<bool> = repeat(false).take(4).collect(); let mut i = 0; let mut color_cnt = 1; while i < n { if colors[i as usize] == 0 { return false; } if graph[i as usize].len() <= 1 { // i號(hào)點(diǎn)是葉節(jié)點(diǎn) i += 1; color_cnt = 1; continue; } has_colors[colors[i as usize] as usize] = true; for near in graph[i as usize].iter() { if !has_colors[colors[*near as usize] as usize] { has_colors[colors[*near as usize] as usize] = true; color_cnt += 1; } } if color_cnt != 3 { return false; } has_colors = repeat(false).take(4).collect(); i += 1; color_cnt = 1; } return true;}
執(zhí)行結(jié)果如下：
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[左神java代碼](github/algorithmzuo/weekly-problems/blob/main/src/class_2022_08_2_week/Code02_TreeDye.java)
