2022-09-27：给定一个棵树， 树上每个节点都有自己的值，记录在数组 nums 里， 比如 nums[4] = 10，表示 4 号点的值是 10， 给定树上的每一条边，记录在二维数组 edges 里， 比如 ed

use std::iter::repeat;fn main() {    let mut nums = vec![1, 5, 5, 4, 11];    let mut edges = vec![vec![0, 1], vec![1, 2], vec![1, 3], vec![3, 4]];    let ans = minimum_score(&mut nums, &mut edges);    println!("ans = {}", ans);}const MAX_VALUE: i32 = 1 << 31 - 1;static mut cnt: i32 = 0;
fn minimum_score(nums: &mut Vec<i32>, edges: &mut Vec<Vec<i32>>) -> i32 {    let n = nums.len() as i32;    // 先建立图    // ArrayList<ArrayList<Integer>> graph = new ArrayList<>();    let mut graph: Vec<Vec<i32>> = vec![];    // 4个点，0、1、2、3    // 0 : {}    // 1 : {}    // 2 : {}    // 3 : {}    for _i in 0..n {        graph.push(vec![]);    }    for edge in edges.iter() {        // a,b        // graph.get(a).add(b);        // graph.get(b).add(a);        graph[edge[0] as usize].push(edge[1]);        graph[edge[1] as usize].push(edge[0]);    }    // 无向边组成的无环图    // 为了方便，就认为0是头    // dfn[i] = ?    let mut dfn: Vec<i32> = repeat(0).take(n as usize).collect();    // xor[i] 以i为头的整棵树，整体异或的结果是多少？    let mut xor: Vec<i32> = repeat(0).take(n as usize).collect();    // size[i] 以i为头的整棵树，一共几个点？    let mut size: Vec<i32> = repeat(0).take(n as usize).collect();
    unsafe {        cnt = 1;    }    dfs(nums, &mut graph, 0, &mut dfn, &mut xor, &mut size);    let mut ans = MAX_VALUE;    let m = edges.len() as i32;    let mut cut1: i32;    let mut cut2: i32;    let mut pre: i32;    let mut pos: i32;    let mut part1: i32;    let mut part2: i32;    let mut part3: i32;    let mut max: i32;    let mut min: i32;    for i in 0..m {        // i，要删掉的第一条边，i号边        // edges[i][0]   edges[i][1]  dfn 谁大，谁就是删掉之后的树的头！cut1        //      a            b                cut1        // { a, b}        //   0  1        let a = edges[i as usize][0];        let b = edges[i as usize][1];        cut1 = if dfn[a as usize] < dfn[b as usize] {            b        } else {            a        };        for j in (i + 1)..m {            // j, 要删掉的第二条边，j号边            // { c, d}            //   0  1            let c = edges[j as usize][0];            let d = edges[j as usize][1];            cut2 = if dfn[c as usize] < dfn[d as usize] {                d            } else {                c            };            // cut1，cut2            pre = if dfn[cut1 as usize] < dfn[cut2 as usize] {                cut1            } else {                cut2            };            pos = if pre == cut1 { cut2 } else { cut1 };            // 早 pre  晚 pos            part1 = xor[pos as usize];            // pos为头的树，是pre为头的树的子树！            if dfn[pos as usize] < dfn[pre as usize] + size[pre as usize] {                part2 = xor[pre as usize] ^ xor[pos as usize];                part3 = xor[0] ^ xor[pre as usize];            } else {                // pos为头的树，不是pre为头的树的子树！                part2 = xor[pre as usize];                part3 = xor[0] ^ part1 ^ part2;            }            max = get_max(get_max(part1, part2), part3);            min = get_min(get_min(part1, part2), part3);            ans = get_min(ans, max - min);        }    }    return ans;}
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    }}// 所有节点的值，存在nums数组里// 整个图结构，存在graph里// 当前来到的是cur号点// 请把cur为头，整棵树，所有节点的dfn、size、xor填好！// 返回！fn dfs(    nums: &mut Vec<i32>,    graph: &mut Vec<Vec<i32>>,    cur: i32,    dfn: &mut Vec<i32>,    xor: &mut Vec<i32>,    size: &mut Vec<i32>,) {    // 当前节点了！，    dfn[cur as usize] = unsafe { cnt };    unsafe {        cnt += 1;    }    // 只是来到了cur的头部！    xor[cur as usize] = nums[cur as usize];    size[cur as usize] = 1;    // 遍历所有的孩子！    for next in graph.clone()[cur as usize].iter() {        //有clone，会影响性能        // 只有dfn是0的孩子，才是cur在树中的下级！！！！        if dfn[*next as usize] == 0 {            // cur某个孩子是next
            dfs(nums, graph, *next, dfn, xor, size);            // next整棵树的异或和，            xor[cur as usize] ^= xor[*next as usize];            // next整棵树的size            size[cur as usize] += size[*next as usize];        }    }}