aboutsummaryrefslogtreecommitdiff
path: root/other/divide_nodes_max_groups.cpp
diff options
context:
space:
mode:
Diffstat (limited to 'other/divide_nodes_max_groups.cpp')
-rw-r--r--other/divide_nodes_max_groups.cpp75
1 files changed, 0 insertions, 75 deletions
diff --git a/other/divide_nodes_max_groups.cpp b/other/divide_nodes_max_groups.cpp
deleted file mode 100644
index ea07a34..0000000
--- a/other/divide_nodes_max_groups.cpp
+++ /dev/null
@@ -1,75 +0,0 @@
-// 2493. Divide Nodes Into the Maximum Number of Groups
-#include "leetcode.h"
-
-/*
- * you are given a positive integer 'n' representing the number of nodes in an
- * undirected graph. the nodes are labeled from 1 to 'n'. you are also given a
- * 2d integer array 'edges', where 'edges[i] = [a_i, b_i]' indicates that there
- * is a bidirectional edge between nodes 'a_i' and 'b_i'. divide the nodes of
- * the graph into 'm' groups such that each node in the graph belongs to exactly
- * one group, and for every pair of nodes in the graph that are connected by an
- * edge '[a_i, b_i]', then '|y - x| = 1'. return the maximum number of groups
- * into which you can divide the nodes. return -1 if it is impossible.
- */
-
-class Solution {
-public:
- int magnificentSets(int n, vector<vector<int>> &edges) {
- vector<int> vis(n);
- vector<vector<int>> graph(n), group;
- for (auto e : edges) {
- graph[e[0] - 1].push_back(e[1] - 1);
- graph[e[1] - 1].push_back(e[0] - 1);
- }
- for (int i = 0; i < n; ++i)
- if (!vis[i]) {
- vis[i] = true;
- stack<int> stk;
- stk.push(i);
- group.push_back({i});
- while (stk.size()) {
- auto u = stk.top();
- stk.pop();
- for (auto &v : graph[u])
- if (!vis[v]) {
- vis[v] = vis[u] + 1;
- stk.push(v);
- group.back().push_back(v);
- } else if ((vis[u] & 1) == (vis[v] & 1))
- return -1;
- }
- }
- auto bfs = [&](int x) {
- int res = 0;
- vector<bool> vis(n);
- vis[x] = true;
- queue<int> q;
- q.push(x);
- for (; q.size(); ++res)
- for (int size = q.size(); size; --size) {
- auto u = q.front();
- q.pop();
- for (auto &v : graph[u])
- if (!vis[v]) {
- vis[v] = true;
- q.push(v);
- }
- }
- return res;
- };
- int ans = 0;
- for (auto &g : group) {
- transform(g.begin(), g.end(), g.begin(), bfs);
- ans += *max_element(g.begin(), g.end());
- }
- return ans;
- }
-};
-
-int main() {
- Solution obj;
- vvd e1 = {{1, 2}, {1, 4}, {1, 5}, {2, 6}, {2, 3}, {4, 6}};
- vvd e2 = {{1, 2}, {2, 3}, {3, 1}};
- printf("%d\n", obj.magnificentSets(6, e1)); // expect: 4
- printf("%d\n", obj.magnificentSets(3, e2)); // expect: -1
-}