From e93cb2c0e651241831aa92eb8891cfae9a3666eb Mon Sep 17 00:00:00 2001
From: gameloader <ggwqqo@163.com>
Date: Thu, 30 Jan 2025 12:18:23 +0800
Subject: [PATCH] leetcode update

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 content/posts/leetcode.md | 118 ++++++++++++++++++++++++++++++++++++++
 1 file changed, 118 insertions(+)

diff --git a/content/posts/leetcode.md b/content/posts/leetcode.md
index 2d8eced..67038a0 100644
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@@ -22374,3 +22374,121 @@ public:
     }
 };
 ```
+## day327 2025-01-30
+### 2493. Divide Nodes Into the Maximum Number of Groups 
+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] = [ai, bi] indicates that there is a bidirectional edge between nodes ai and bi. Notice that the given graph may be disconnected.
+
+Divide the nodes of the graph into m groups (1-indexed) such that:
+
+Each node in the graph belongs to exactly one group.
+For every pair of nodes in the graph that are connected by an edge [ai, bi], if ai belongs to the group with index x, and bi belongs to the group with index y, then |y - x| = 1.
+Return the maximum number of groups (i.e., maximum m) into which you can divide the nodes. Return -1 if it is impossible to group the nodes with the given conditions.
+
+![0130utjkuecb0rBs](https://testingcf.jsdelivr.net/gh/game-loader/picbase@master/uPic/0130utjkuecb0rBs.png)
+
+### 题解
+本题是一道难题，也是看了题解才明白如何解决该题。
+
+要想解决该题，首先要观察到一个非常重要的现象，即任何可以被分成两个以上不同组的连通图中的所有节点，最终都可以通过合并归类到两个组中。如当前有四个不同的组，则一定可以将第四组和第二组合并到一起，因为二者都和第三组相连，二第四组和第一组又无关，因此合并后一定满足条件，现在变为三组，同理可以将第三组和第一组合并，这样就变成了两组，由此可以发现，只要图能够被分为两组，且只有位于不同组的节点之间有边，该图就可以被分成不同的组，即图必须是一个二分图。
+
+判断二分图的常用办法即涂色法，即将图中节点分为两组后，每组中的全部节点都是同一种颜色。在实际实现时，只需将每个节点的相邻节点都涂为和当前节点不同的颜色即可，一旦出现冲突即说明该图不是二分图。
+
+第二个难点是观察到是二分图的情况下，一个连通图最多可以被分成的组的个数即为图的直径(图中两个节点之间最远的距离)，其实想到了之后验证起来就会觉得这也是一个很显然的事情（会的不难，难的不会，验证想法的难度总小于想到想法本身），只考虑直径这条路径，则这条路径从任意一端开始，执行bfs，将一次bfs中同一层的所有节点都放在同一个组中，如此反复，最终到直径的另一端，一定可以得到一种有效的分组，而直径又是图中的两节点之间的最长距离，因此不可能有数量更多的分组了。无向图的直径的计算最简便直接的方法是对每个节点都执行bfs，找到离该节点最远的节点的距离，所有距离中的最大值即为该图的直径。
+
+### 代码
+```cpp 
+class Solution {
+public:
+    int bfs(int start, vector<vector<int>>& adj, int n) {
+        vector<int> levels(n + 1, -1);
+        queue<int> q;
+        q.push(start);
+        levels[start] = 0;
+        int maxLevel = 0;
+        
+        while (!q.empty()) {
+            int curr = q.front();
+            q.pop();
+            
+            for (int next : adj[curr]) {
+                if (levels[next] == -1) {
+                    levels[next] = levels[curr] + 1;
+                    maxLevel = max(maxLevel, levels[next]);
+                    q.push(next);
+                }
+            }
+        }
+        return maxLevel + 1; // 返回最大可能的分组数
+    }
+    
+    bool isBipartite(int start, vector<vector<int>>& adj, vector<int>& color, int n) {
+        queue<int> q;
+        q.push(start);
+        color[start] = 0;
+        
+        while (!q.empty()) {
+            int curr = q.front();
+            q.pop();
+            
+            for (int next : adj[curr]) {
+                if (color[next] == -1) {
+                    color[next] = 1 - color[curr];
+                    q.push(next);
+                } else if (color[next] == color[curr]) {
+                    return false;
+                }
+            }
+        }
+        return true;
+    }
+    
+    void dfs(int node, vector<vector<int>>& adj, vector<bool>& visited, vector<int>& component) {
+        visited[node] = true;
+        component.push_back(node);
+        
+        for (int next : adj[node]) {
+            if (!visited[next]) {
+                dfs(next, adj, visited, component);
+            }
+        }
+    }
+    
+    int magnificentSets(int n, vector<vector<int>>& edges) {
+        vector<vector<int>> adj(n + 1);
+        for (const auto& edge : edges) {
+            adj[edge[0]].push_back(edge[1]);
+            adj[edge[1]].push_back(edge[0]);
+        }
+        
+        vector<bool> visited(n + 1, false);
+        vector<vector<int>> components;
+        
+        for (int i = 1; i <= n; i++) {
+            if (!visited[i]) {
+                vector<int> component;
+                dfs(i, adj, visited, component);
+                components.push_back(component);
+            }
+        }
+        
+        int result = 0;
+        for (const auto& component : components) {
+            vector<int> color(n + 1, -1);
+            if (!isBipartite(component[0], adj, color, n)) {
+                return -1;
+            }
+            
+            // 对于每个连通分量，尝试从每个节点开始BFS，取最大值
+            int maxGroups = 0;
+            for (int node : component) {
+                maxGroups = max(maxGroups, bfs(node, adj, n));
+            }
+            result += maxGroups;
+        }
+        
+        return result;
+    }
+};
+```