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Commit d877391

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第454场周赛T1~T4 (4)
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public class Solution3582 {
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public String generateTag(String caption) {
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caption = ("#" + caption.strip()).toLowerCase();
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StringBuilder ans = new StringBuilder("#");
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for (int i = 1; i < caption.length(); i++) {
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char c = caption.charAt(i);
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if (c == ' ') continue;
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if (caption.charAt(i - 1) == ' ') {
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ans.append(Character.toUpperCase(c));
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} else {
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ans.append(c);
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}
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}
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return ans.substring(0, Math.min(100, ans.length()));
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}
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}
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/*
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3582. 为视频标题生成标签
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https://leetcode.cn/problems/generate-tag-for-video-caption/description/
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第 454 场周赛 T1。
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给你一个字符串 caption,表示一个视频的标题。
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需要按照以下步骤 按顺序 生成一个视频的 有效标签 :
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1.将 所有单词 组合为单个 驼峰命名字符串 ,并在前面加上 '#'。驼峰命名字符串 指的是除第一个单词外,其余单词的首字母大写,且每个单词的首字母之后的字符必须是小写。
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2.移除 所有不是英文字母的字符,但 保留 第一个字符 '#'。
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3.将结果 截断 为最多 100 个字符。
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对 caption 执行上述操作后,返回生成的 标签 。
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提示:
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1 <= caption.length <= 150
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caption 仅由英文字母和 ' ' 组成。
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模拟。WA了5次,细节是魔鬼。
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时间复杂度 O(n)。
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*/
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import java.util.HashMap;
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import java.util.Map;
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public class Solution3583 {
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private static final int MOD = (int) (1e9 + 7);
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public int specialTriplets(int[] nums) {
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int n = nums.length;
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int[] lcnt = new int[n];
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Map<Integer, Integer> cnt = new HashMap<>();
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for (int i = 0; i < n; i++) {
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lcnt[i] = cnt.getOrDefault(nums[i] * 2, 0);
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cnt.merge(nums[i], 1, Integer::sum);
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}
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int[] rcnt = new int[n];
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cnt.clear();
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for (int i = n - 1; i >= 0; i--) {
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rcnt[i] = cnt.getOrDefault(nums[i] * 2, 0);
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cnt.merge(nums[i], 1, Integer::sum);
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}
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long ans = 0;
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for (int i = 1; i + 1 < n; i++) {
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ans = (ans + (long) lcnt[i] * rcnt[i]) % MOD;
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}
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return (int) ans;
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}
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}
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/*
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3583. 统计特殊三元组
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https://leetcode.cn/problems/count-special-triplets/description/
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第 454 场周赛 T2。
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给你一个整数数组 nums。
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特殊三元组 定义为满足以下条件的下标三元组 (i, j, k):
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- 0 <= i < j < k < n,其中 n = nums.length
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- nums[i] == nums[j] * 2
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- nums[k] == nums[j] * 2
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返回数组中 特殊三元组 的总数。
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由于答案可能非常大,请返回结果对 10^9 + 7 取余数后的值。
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提示:
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3 <= n == nums.length <= 10^5
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0 <= nums[i] <= 10^5
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前后缀分解 + 枚举中间。
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时间复杂度 O(n)。
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*/
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public class Solution3584 {
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private static final long INF = (long) 1e18;
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public long maximumProduct(int[] nums, int m) {
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int n = nums.length;
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if (m == 1) {
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long ans = 0;
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for (int x : nums) ans = Math.max(ans, (long) x * x);
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return ans;
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}
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int k = m - 1;
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int[] max_suffix = new int[n];
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int[] min_suffix = new int[n];
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max_suffix[n - 1] = nums[n - 1];
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min_suffix[n - 1] = nums[n - 1];
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for (int i = n - 2; i >= 0; i--) {
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max_suffix[i] = Math.max(nums[i], max_suffix[i + 1]);
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min_suffix[i] = Math.min(nums[i], min_suffix[i + 1]);
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}
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long candidate;
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long max_prod = -INF;
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for (int i = 0; i + k < n; i++) {
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int start = i + k;
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if (nums[i] >= 0) {
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candidate = (long) nums[i] * max_suffix[start];
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} else {
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candidate = (long) nums[i] * min_suffix[start];
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}
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max_prod = Math.max(max_prod, candidate);
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}
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return max_prod;
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}
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}
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/*
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3584. 子序列首尾元素的最大乘积
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https://leetcode.cn/problems/maximum-product-of-first-and-last-elements-of-a-subsequence/description/
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第 454 场周赛 T3。
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给你一个整数数组 nums 和一个整数 m。
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返回任意大小为 m 的 子序列 中首尾元素乘积的最大值。
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子序列 是可以通过删除原数组中的一些元素(或不删除任何元素),且不改变剩余元素顺序而得到的数组。
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提示:
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1 <= nums.length <= 10^5
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-10^5 <= nums[i] <= 10^5
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1 <= m <= nums.length
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预处理后缀 + 贪心。
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时间复杂度 O(n)。
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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 Solution3585 {
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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 long[] sum_to_root;
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public int[] findMedian(int n, int[][] edges, int[][] queries) {
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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 long[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 u = queries[i][0], v = queries[i][1];
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long total = get_path_sum(u, v);
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int lcaNode = getLCA(u, v);
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int current1 = u;
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for (int k = mx - 1; k >= 0; k--) {
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int next_node = pa[current1][k];
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if (next_node == -1) continue;
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if (dep[next_node] < dep[lcaNode]) continue;
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if (2 * (sum_to_root[u] - sum_to_root[next_node]) < total) {
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current1 = next_node;
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}
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}
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if (current1 != lcaNode) {
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ans[i] = pa[current1][0];
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} else {
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int candidate2 = v;
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for (int k = mx - 1; k >= 0; k--) {
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int next_node = pa[candidate2][k];
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if (next_node == -1) continue;
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if (dep[next_node] <= dep[lcaNode]) continue;
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if (2 * (sum_to_root[u] + sum_to_root[next_node] - 2 * sum_to_root[lcaNode]) >= total) {
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candidate2 = next_node;
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}
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}
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ans[i] = candidate2;
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}
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}
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return ans;
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}
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long 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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3585. 树中找到带权中位节点
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https://leetcode.cn/problems/find-weighted-median-node-in-tree/description/
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第 454 场周赛 T4。
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给你一个整数 n,以及一棵 无向带权 树,根节点为节点 0,树中共有 n 个节点,编号从 0 到 n - 1。该树由一个长度为 n - 1 的二维数组 edges 表示,其中 edges[i] = [ui, vi, wi] 表示存在一条从节点 ui 到 vi 的边,权重为 wi。
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带权中位节点 定义为从 ui 到 vi 路径上的 第一个 节点 x,使得从 ui 到 x 的边权之和 大于等于 该路径总权值和的一半。
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给你一个二维整数数组 queries。对于每个 queries[j] = [uj, vj],求出从 uj 到 vj 路径上的带权中位节点。
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返回一个数组 ans,其中 ans[j] 表示查询 queries[j] 的带权中位节点编号。
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提示:
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2 <= n <= 10^5
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edges.length == n - 1
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edges[i] == [ui, vi, wi]
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0 <= ui, vi < n
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1 <= wi <= 10^9
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1 <= queries.length <= 10^5
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queries[j] == [uj, vj]
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0 <= uj, vj < n
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输入保证 edges 表示一棵合法的树。
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最近公共祖先 LCA + 树上倍增。
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先从 u 点倍增到 lca 点,若干未能得到答案,那么再从 lca 点倍增到 v 点(反向)。
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时间复杂度 O((n+q)logn)。
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rating 2399 (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 Solution3582Tests {
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private final Solution3582 solution3582 = new Solution3582();
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@Test
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public void example1() {
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String caption = "Leetcode daily streak achieved";
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String expected = "#leetcodeDailyStreakAchieved";
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Assertions.assertEquals(expected, solution3582.generateTag(caption));
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}
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@Test
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public void example2() {
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String caption = "can I Go There";
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String expected = "#canIGoThere";
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Assertions.assertEquals(expected, solution3582.generateTag(caption));
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}
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@Test
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public void example3() {
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String caption = "hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh";
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String expected = "#hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh";
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Assertions.assertEquals(expected, solution3582.generateTag(caption));
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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 Solution3583Tests {
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private final Solution3583 solution3583 = new Solution3583();
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@Test
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public void example1() {
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int[] nums = {6, 3, 6};
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int expected = 1;
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Assertions.assertEquals(expected, solution3583.specialTriplets(nums));
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}
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@Test
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public void example2() {
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int[] nums = {0, 1, 0, 0};
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int expected = 1;
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Assertions.assertEquals(expected, solution3583.specialTriplets(nums));
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}
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@Test
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public void example3() {
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int[] nums = {8, 4, 2, 8, 4};
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int expected = 2;
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Assertions.assertEquals(expected, solution3583.specialTriplets(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 Solution3584Tests {
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private final Solution3584 solution3584 = new Solution3584();
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@Test
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public void example1() {
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int[] nums = {-1, -9, 2, 3, -2, -3, 1};
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int m = 1;
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long expected = 81;
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Assertions.assertEquals(expected, solution3584.maximumProduct(nums, m));
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}
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@Test
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public void example2() {
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int[] nums = {1, 3, -5, 5, 6, -4};
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int m = 3;
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long expected = 20;
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Assertions.assertEquals(expected, solution3584.maximumProduct(nums, m));
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}
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@Test
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public void example3() {
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int[] nums = {2, -1, 2, -6, 5, 2, -5, 7};
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int m = 2;
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long expected = 35;
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Assertions.assertEquals(expected, solution3584.maximumProduct(nums, m));
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}
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}

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