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2846. Minimum Edge Weight Equilibrium Queries in a Tree
类似题目: 1483. Kth Ancestor of a Tree Node
LCA模版 【模板讲解】树上倍增算法(以及最近公共祖先) LCA 模板(Python/Java/C++/Go)
class Solution { public int[] minOperationsQueries(int n, int[][] edges, int[][] queries) { int queriesLen = queries.length; int[] ans = new int[queriesLen]; Map<Integer, Set<int[]>> g = new HashMap<>(); for (int[] edge : edges) { int u = edge[0], v = edge[1], w = edge[2]; g.putIfAbsent(u, new HashSet<>()); g.putIfAbsent(v, new HashSet<>()); g.get(u).add(new int[]{v, w}); g.get(v).add(new int[]{u, w}); } int m = 32 - Integer.numberOfLeadingZeros(n); int[][] pa = new int[n][m]; for (int i = 0; i < n; i++) { Arrays.fill(pa[i], -1); } int[][][] cnt = new int[n][m][26]; int[] depth = new int[n]; dfs(0, -1, g, pa, cnt, depth); // init for (int j = 0; j < m - 1; j++) { for (int i = 0; i < n; i++) { int fa = pa[i][j]; if (fa != -1) { pa[i][j + 1] = pa[fa][j]; for (int k = 0; k < 26; k++) { cnt[i][j + 1][k] += cnt[i][j][k] + cnt[fa][j][k]; } } } } // lca for (int i = 0; i < queriesLen; i++) { int x = queries[i][0], y = queries[i][1]; if (depth[x] > depth[y]) { int tmp = x; x = y; y = tmp; } int pathLen = depth[y] + depth[x]; int[] cntDiff = new int[26]; // put x and y at same level for (int j = depth[y] - depth[x]; j > 0; j &= (j - 1)) { int k = Integer.numberOfTrailingZeros(j); for (int l = 0; l < 26; l++) { cntDiff[l] += cnt[y][k][l]; } y = pa[y][k]; } if (x != y) { for (int j = m - 1; j >= 0; j--) { int px = pa[x][j], py = pa[y][j]; if (px != py) { for (int k = 0; k < 26; k++) { cntDiff[k] += cnt[x][j][k] + cnt[y][j][k]; } x = px; y = py; } } for (int k = 0; k < 26; k++) { cntDiff[k] += cnt[x][0][k] + cnt[y][0][k]; } x = pa[x][0]; } int lca = x; pathLen -= 2 * depth[lca]; int max = 0; for (int j = 0; j < 26; j++) { max = Math.max(max, cntDiff[j]); } ans[i] = pathLen - max; } return ans; } private void dfs(int u, int p, Map<Integer, Set<int[]>> g, int[][] pa, int[][][] cnt, int[] depth) { pa[u][0] = p; for (int[] edge : g.getOrDefault(u, new HashSet<>())) { int v = edge[0], w = edge[1]; if (v == p) continue; depth[v] = depth[u] + 1; cnt[v][0][w - 1]++; dfs(v, u, g, pa, cnt, depth); } } }
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2846. Minimum Edge Weight Equilibrium Queries in a Tree
2846. Minimum Edge Weight Equilibrium Queries in a Tree
类似题目:
1483. Kth Ancestor of a Tree Node
LCA模版
【模板讲解】树上倍增算法(以及最近公共祖先)
LCA 模板(Python/Java/C++/Go)
The text was updated successfully, but these errors were encountered: