Minimum Edge Weight Equilibrium Queries in a Tree
Minimum Edge Weight Equilibrium Queries in a Tree
There is an undirected tree with n nodes labeled from 0 to n - 1. You are given the integer n and a 2D integer array edges of length n - 1, where edges[i] = [ui, vi, wi] indicates that there is an edge between nodes ui and vi with weight wi in the tree.
You are also given a 2D integer array queries of length m, where queries[i] = [ai, bi]. For each query, find the minimum number of operations required to make the weight of every edge on the path from ai to bi equal. In one operation, you can choose any edge of the tree and change its weight to any value.
Note that:
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The path from ai to bi is a sequence of distinct nodes starting with node ai and ending with node bi
Return an array answer of length m where answer[i] is the answer to the ith
Input: n = 7, edges = [[0,1,1],[1,2,1],[2,3,1],[3,4,2],[4,5,2],[5,6,2]], queries = [[0,3],[3,6],[2,6],[0,6]] Output: [0,0,1,3] Explanation: In the first query, all the edges in the path from 0 to 3 have a weight of 1. Hence, the answer is 0. In the second query, all the edges in the path from 3 to 6 have a weight of 2. Hence, the answer is 0. In the third query, we change the weight of edge [2,3] to 2. After this operation, all the edges in the path from 2 to 6 have a weight of 2. Hence, the answer is 1. In the fourth query, we change the weights of edges [0,1], [1,2] and [2,3] to 2. After these operations, all the edges in the path from 0 to 6 have a weight of 2. Hence, the answer is 3. For each queries[i], it can be shown that answer[i] is the minimum number of operations needed to equalize all the edge weights in the path from ai to bi.
Example 2:
Input: n = 8, edges = [[1,2,6],[1,3,4],[2,4,6],[2,5,3],[3,6,6],[3,0,8],[7,0,2]], queries = [[4,6],[0,4],[6,5],[7,4]] Output: [1,2,2,3] Explanation: In the first query, we change the weight of edge [1,3] to 6. After this operation, all the edges in the path from 4 to 6 have a weight of 6. Hence, the answer is 1. In the second query, we change the weight of edges [0,3] and [3,1] to 6. After these operations, all the edges in the path from 0 to 4 have a weight of 6. Hence, the answer is 2. In the third query, we change the weight of edges [1,3] and [5,2] to 6. After these operations, all the edges in the path from 6 to 5 have a weight of 6. Hence, the answer is 2. In the fourth query, we change the weights of edges [0,7], [0,3] and [1,3] to 6. After these operations, all the edges in the path from 7 to 4 have a weight of 6. Hence, the answer is 3. For each queries[i], it can be shown that answer[i] is the minimum number of operations needed to equalize all the edge weights in the path from ai to bi.
Constraints:
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1 <= n <= 104
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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 <= 26
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The input is generated such that edges represents a valid tree.
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1 <= queries.length == m <= 2 * 104
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queries[i].length == 2
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0 <= ai, bi < n
解题思路
周赛很顺利做出这题,上大分,写篇题解纪念一下。
首先看到边的权值最大才,因此很容易想到对于每个询问直接暴力枚举把边都变成哪个权值,然后再取最小的操作次数。关键是要知道两个节点所构成的简单路径中每种权值的边共有多少。
假定树的根节点是号点,直接暴力枚举所有可能的边权,分别统计从号节点出发到其他点的路径中有多少条权值恰好为的边。令表示从节点到节点的路径中权值为的边的数量,因此如果要将路径中的所有边的权值变成,那么最小需要的操作次数就是,其中,表示从节点到节点的路径长度。
所以对于每个询问直接暴力枚举所有可能的,然后取上式的最小值即可,即
AC代码如下,时间复杂度为:
1 class Solution { 2 public: 3 vector<int> minOperationsQueries(int n, vector<vector<int>>& edges, vector<vector<int>>& queries) { 4 vector<vector<vector<int>>> g(n + 1); 5 for (auto &p : edges) { 6 p[0]++, p[1]++; 7 g[p[0]].push_back({p[1], p[2]}); 8 g[p[1]].push_back({p[0], p[2]}); 9 } 10 vector<vector<int>> fa(n + 1, vector<int>(14)); 11 vector<int> dep(n + 1, 0x3f3f3f3f); 12 dep[0] = 0, dep[1] = 1; 13 queue<int> q({1}); 14 while (!q.empty()) { 15 int t = q.front(); 16 q.pop(); 17 for (auto &p : g[t]) { 18 if (dep[p[0]] > dep[t] + 1) { 19 dep[p[0]] = dep[t] + 1; 20 q.push(p[0]); 21 fa[p[0]][0] = t; 22 for (int i = 1; i <= 13; i++) { 23 fa[p[0]][i] = fa[fa[p[0]][i - 1]][i - 1]; 24 } 25 } 26 } 27 } 28 vector<vector<int>> s(27, vector<int>(n + 1)); 29 for (int w = 1; w <= 26; w++) { 30 function<void(int, int)> dfs = [&](int u, int pre) { 31 for (auto &p : g[u]) { 32 if (p[0] != pre) { 33 s[w][p[0]] = s[w][u] + (p[1] == w); 34 dfs(p[0], u); 35 } 36 } 37 }; 38 dfs(1, -1); 39 } 40 vector<int> ans; 41 function<int(int, int)> lca = [&](int a, int b) { 42 if (dep[a] < dep[b]) swap(a, b); 43 for (int i = 13; i >= 0; i--) { 44 if (dep[fa[a][i]] >= dep[b]) a = fa[a][i]; 45 } 46 if (a == b) return a; 47 for (int i = 13; i >= 0; i--) { 48 if (fa[a][i] != fa[b][i]) a = fa[a][i], b = fa[b][i]; 49 } 50 return fa[a][0]; 51 }; 52 for (auto &p : queries) { 53 p[0]++, p[1]++; 54 int x = lca(p[0], p[1]), t = n; 55 for (int w = 1; w <= 26; w++) { 56 t = min(t, dep[p[0]] + dep[p[1]] - 2 * dep[x] - (s[w][p[0]] + s[w][p[1]] - 2 * s[w][x])); 57 } 58 ans.push_back(t); 59 } 60 return ans; 61 } 62 };
本文来自博客园,作者:onlyblues,转载请注明原文链接:https://www.cnblogs.com/onlyblues/p/17677305.html
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