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Commit 3b6b5c6

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第450场周赛T1~T4 (4)
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public class Solution3550 {
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public int smallestIndex(int[] nums) {
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int n = nums.length;
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for (int i = 0; i < n; i++) {
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if (digitSum(nums[i]) == i) {
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return i;
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}
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}
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return -1;
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}
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private int digitSum(int x) {
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int sum = 0;
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while (x > 0) {
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sum += x % 10;
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x /= 10;
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}
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return sum;
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}
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}
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/*
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3550. 数位和等于下标的最小下标
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https://leetcode.cn/problems/smallest-index-with-digit-sum-equal-to-index/description/
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第 450 场周赛 T1。
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给你一个整数数组 nums 。
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返回满足 nums[i] 的数位和(每一位数字相加求和)等于 i 的 最小 下标 i 。
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如果不存在满足要求的下标,返回 -1 。
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提示:
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1 <= nums.length <= 100
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0 <= nums[i] <= 1000
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模拟 遍历。
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*/
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import java.util.Arrays;
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import java.util.Comparator;
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public class Solution3551 {
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public int minSwaps(int[] nums) {
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int n = nums.length;
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int[] b = new int[n]; // 数位和
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for (int i = 0; i < n; i++) {
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b[i] = digitSum(nums[i]);
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}
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// 置换环:对于每个环,交换次数为环的大小减一。
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Integer[] ids = new Integer[n]; // 将 a 离散化
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for (int i = 0; i < n; i++) ids[i] = i;
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// 如果两个数字的数位和相等,则较小的数字排在前面。
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Arrays.sort(ids, Comparator.comparingInt(o -> b[(int) o])
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.thenComparingInt(o -> nums[(int) o]));
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int ans = n;
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boolean[] vis = new boolean[n];
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for (Integer v : ids) {
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if (vis[v]) continue;
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while (!vis[v]) {
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vis[v] = true;
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v = ids[v];
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}
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ans -= 1;
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}
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return ans;
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}
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private int digitSum(int x) {
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int sum = 0;
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while (x > 0) {
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sum += x % 10;
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x /= 10;
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}
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return sum;
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}
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}
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/*
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3551. 数位和排序需要的最小交换次数
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https://leetcode.cn/problems/minimum-swaps-to-sort-by-digit-sum/description/
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第 450 场周赛 T2。
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给你一个由 互不相同 的正整数组成的数组 nums,需要根据每个数字的数位和(即每一位数字相加求和)按 升序 对数组进行排序。如果两个数字的数位和相等,则较小的数字排在前面。
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返回将 nums 排列为上述排序顺序所需的 最小 交换次数。
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一次 交换 定义为交换数组中两个不同位置的值。
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提示:
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1 <= nums.length <= 10^5
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1 <= nums[i] <= 10^9
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nums 由 互不相同 的正整数组成。
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置换环。
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相似题目: 2471. 逐层排序二叉树所需的最少操作数目
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https://leetcode.cn/problems/minimum-number-of-operations-to-sort-a-binary-tree-by-level/
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*/
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import java.util.ArrayDeque;
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import java.util.ArrayList;
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import java.util.Arrays;
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import java.util.Deque;
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import java.util.List;
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public class Solution3552 {
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private static final int[][] DIRECTIONS = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
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private int[] dist;
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private Deque<int[]> dq;
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public int minMoves(String[] matrix) {
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int m = matrix.length;
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int n = matrix[0].length();
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List<Integer>[] trans = new ArrayList[26];
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Arrays.setAll(trans, e -> new ArrayList<>());
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for (int i = 0; i < m; i++) {
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for (int j = 0; j < n; j++) {
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char c = matrix[i].charAt(j);
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if ('A' <= c && c <= 'Z') {
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trans[c - 'A'].add(i * n + j);
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}
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}
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}
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// 节点编号:格子从 0 到 nm - 1,传送中心从 nm 到 nm + 25
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dist = new int[n * m + 26];
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Arrays.fill(dist, -1);
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dq = new ArrayDeque<>(); // dis, sn
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dq.add(new int[]{0, 0});
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while (!dq.isEmpty()) {
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int[] p = dq.removeFirst();
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int dis = p[0], sn = p[1];
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if (dist[sn] >= 0) continue;
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dist[sn] = dis;
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if (sn < m * n) {
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// 在格子上
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int cx = sn / n;
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int cy = sn % n;
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for (int[] d : DIRECTIONS) {
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int nx = cx + d[0];
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int ny = cy + d[1];
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if (nx >= 0 && nx < m && ny >= 0 && ny < n && matrix[nx].charAt(ny) != '#') {
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addDeque(sn, nx * n + ny, 1);
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}
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}
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// 如果格子上有字母,还可以走到传送中心
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char c = matrix[cx].charAt(cy);
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if ('A' <= c && c <= 'Z') {
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addDeque(sn, n * m + (c - 'A'), 0);
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}
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} else {
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// 在传送中心里
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for (int next_sn : trans[sn - n * m]) {
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addDeque(sn, next_sn, 0);
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}
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}
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}
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return dist[n * m - 1];
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}
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// val:增量 0/1
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private void addDeque(int sn, int next_sn, int val) {
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if (dist[next_sn] >= 0) return;
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if (val == 0) dq.addFirst(new int[]{dist[sn], next_sn});
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else dq.addLast(new int[]{dist[sn] + 1, next_sn});
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}
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}
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/*
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3552. 网格传送门旅游
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https://leetcode.cn/problems/grid-teleportation-traversal/description/
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第 450 场周赛 T3。
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给你一个大小为 m x n 的二维字符网格 matrix,用字符串数组表示,其中 matrix[i][j] 表示第 i 行和第 j 列处的单元格。每个单元格可以是以下几种字符之一:
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- '.' 表示一个空单元格。
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- '#' 表示一个障碍物。
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- 一个大写字母('A' 到 'Z')表示一个传送门。
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你从左上角单元格 (0, 0) 出发,目标是到达右下角单元格 (m - 1, n - 1)。你可以从当前位置移动到相邻的单元格(上、下、左、右),移动后的单元格必须在网格边界内且不是障碍物。
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如果你踏入一个包含传送门字母的单元格,并且你之前没有使用过该传送门字母,你可以立即传送到网格中另一个具有相同字母的单元格。这次传送不计入移动次数,但每个字母对应的传送门在旅程中 最多 只能使用一次。
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返回到达右下角单元格所需的 最少 移动次数。如果无法到达目的地,则返回 -1。
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提示:
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1 <= m == matrix.length <= 10^3
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1 <= n == matrix[i].length <= 10^3
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matrix[i][j] 是 '#'、'.' 或一个大写英文字母。
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matrix[0][0] 不是障碍物。
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0-1 BFS。
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时间复杂度 O(mn)。
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rating 2009 (clist.by)
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*/
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import java.util.ArrayList;
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import java.util.Arrays;
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import java.util.List;
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public class Solution3553 {
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private List<int[]>[] g;
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private int dfn;
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private int[][] nodes; // l, r
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private final int mx = 17;
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private int[][] pa;
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private int[] dep;
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private int[] sum_to_root;
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public int[] minimumWeight(int[][] edges, int[][] queries) {
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int n = edges.length + 1;
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g = new ArrayList[n];
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Arrays.setAll(g, e -> new ArrayList<>());
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for (int[] e : edges) {
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g[e[0]].add(new int[]{e[1], e[2]});
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g[e[1]].add(new int[]{e[0], e[2]});
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}
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dfn = 0;
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nodes = new int[n][2];
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pa = new int[n][mx];
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dep = new int[n];
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sum_to_root = new int[n];
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build(0, -1);
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for (int i = 0; i + 1 < mx; i++) {
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for (int v = 0; v < pa.length; v++) {
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int p = pa[v][i];
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if (p != -1) {
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pa[v][i + 1] = pa[p][i];
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} else {
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pa[v][i + 1] = -1;
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}
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}
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}
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int[] ans = new int[queries.length];
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for (int i = 0; i < queries.length; i++) {
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int src1 = queries[i][0], src2 = queries[i][1], dest = queries[i][2];
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int sum1 = get_path_sum(src1, dest);
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int sum2 = get_path_sum(src2, dest);
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int sum3 = get_path_sum(src1, src2);
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ans[i] = (sum1 + sum2 + sum3) / 2;
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}
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return ans;
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}
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int get_path_sum(int u, int v) {
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int ancestor = getLCA(u, v);
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return sum_to_root[u] + sum_to_root[v] - 2 * sum_to_root[ancestor];
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}
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int build(int v, int fa) {
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dfn++;
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nodes[v][0] = dfn;
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pa[v][0] = fa;
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int sz = 1;
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for (int[] p : g[v]) {
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int w = p[0], wt = p[1];
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if (w != fa) {
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dep[w] = dep[v] + 1;
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sum_to_root[w] = sum_to_root[v] + wt;
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sz += build(w, v);
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}
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}
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nodes[v][1] = nodes[v][0] + sz - 1;
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return sz;
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}
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int uptoDep(int v, int d) {
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for (int k = dep[v] - d; k > 0; k &= k - 1) {
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v = pa[v][Integer.numberOfTrailingZeros(k)];
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}
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return v;
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}
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int getLCA(int v, int w) {
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if (dep[v] > dep[w]) {
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int tmp = v;
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v = w;
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w = tmp;
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}
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w = uptoDep(w, dep[v]);
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if (w == v) return v;
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for (int i = mx - 1; i >= 0; i--) {
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if (pa[v][i] != pa[w][i]) {
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v = pa[v][i];
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w = pa[w][i];
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}
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}
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return pa[v][0];
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}
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}
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/*
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3553. 包含给定路径的最小带权子树 II
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https://leetcode.cn/problems/minimum-weighted-subgraph-with-the-required-paths-ii/description/
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第 450 场周赛 T4。
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给你一个 无向带权 树,共有 n 个节点,编号从 0 到 n - 1。这棵树由一个二维整数数组 edges 表示,长度为 n - 1,其中 edges[i] = [ui, vi, wi] 表示存在一条连接节点 ui 和 vi 的边,权重为 wi。
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此外,给你一个二维整数数组 queries,其中 queries[j] = [src1j, src2j, destj]。
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返回一个长度等于 queries.length 的数组 answer,其中 answer[j] 表示一个子树的 最小总权重 ,使用该子树的边可以从 src1j 和 src2j 到达 destj 。
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这里的 子树 是指原树中任意节点和边组成的连通子集形成的一棵有效树。
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提示:
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3 <= n <= 10^5
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edges.length == n - 1
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edges[i].length == 3
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0 <= ui, vi < n
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1 <= wi <= 104
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1 <= queries.length <= 10^5
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queries[j].length == 3
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0 <= src1j, src2j, destj < n
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src1j、src2j 和 destj 互不不同。
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输入数据保证 edges 表示的是一棵有效的树。
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最近公共祖先 LCA。
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时间复杂度 O((n+q)logn)。
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相似题目: E. Jamie and Tree
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https://codeforces.com/contest/916/problem/E
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3559. 给边赋权值的方案数 II
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https://leetcode.cn/problems/number-of-ways-to-assign-edge-weights-ii/description/
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rating 2413 (clist.by)
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*/
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import org.junit.jupiter.api.Assertions;
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import org.junit.jupiter.api.Test;
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public class Solution3550Tests {
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private final Solution3550 solution3550 = new Solution3550();
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@Test
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public void example1() {
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int[] nums = {1, 3, 2};
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int expected = 2;
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Assertions.assertEquals(expected, solution3550.smallestIndex(nums));
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}
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@Test
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public void example2() {
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int[] nums = {1, 10, 11};
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int expected = 1;
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Assertions.assertEquals(expected, solution3550.smallestIndex(nums));
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}
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@Test
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public void example3() {
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int[] nums = {1, 2, 3};
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int expected = -1;
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Assertions.assertEquals(expected, solution3550.smallestIndex(nums));
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}
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}
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import org.junit.jupiter.api.Assertions;
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import org.junit.jupiter.api.Test;
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public class Solution3551Tests {
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private final Solution3551 solution3551 = new Solution3551();
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@Test
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public void example1() {
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int[] nums = {37, 100};
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int expected = 1;
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Assertions.assertEquals(expected, solution3551.minSwaps(nums));
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}
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@Test
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public void example2() {
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int[] nums = {22, 14, 33, 7};
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int expected = 0;
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Assertions.assertEquals(expected, solution3551.minSwaps(nums));
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}
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@Test
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public void example3() {
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int[] nums = {18, 43, 34, 16};
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int expected = 2;
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Assertions.assertEquals(expected, solution3551.minSwaps(nums));
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}
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}
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import org.junit.jupiter.api.Assertions;
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import org.junit.jupiter.api.Test;
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public class Solution3552Tests {
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private final Solution3552 solution3552 = new Solution3552();
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@Test
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public void example1() {
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String[] matrix = {"A..", ".A.", "..."};
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int expected = 2;
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Assertions.assertEquals(expected, solution3552.minMoves(matrix));
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}
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@Test
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public void example2() {
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String[] matrix = {".#...", ".#.#.", ".#.#.", "...#."};
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int expected = 13;
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Assertions.assertEquals(expected, solution3552.minMoves(matrix));
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}
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}

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