LintCode 717: Tree Longest Path With Same Value

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本文介绍了一个算法问题，即在一个带有整数值标签的无向树中寻找具有相同节点值的最长路径，通过深度优先搜索（DFS）策略实现。文章提供了详细的示例和解决方案，包括输入输出示例以及代码实现。

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Tree Longest Path With Same Value
中文English
We consider an undirected tree with N nodes, numbered from 1 to N, Additionally, each node also has a label attached to it and the label is an integer value. Note that different nodes can have identical labels. You need to write a function , that , given a zero-indexed array A of length N, where A[J] is the label value of the (J + 1)-th node in the tree, and a zero-indexed array E of length K = (N - 1) * 2 in which the edges of the tree are described (for every 0 <= j <= N -2 values E[2 * J] and E[2 * J + 1] represents and edge connecting node E[2 * J] with node E[2 * J + 1])， returns the length of the longest path such that all the nodes on that path have the same label. Then length of a path if defined as the number of edges in that path.

Example
Example1

Input: A = [1, 1, 1 ,2, 2] and E = [1, 2, 1, 3, 2, 4, 2, 5]
Output: 2
Explanation:
described tree appears as follows:
1 （value = 1） / \ z (value = 1) 2 3 (value = 1) / \ (value = 2) 4 5 (value = 2)
The longest path (in which all nodes have the save value of 1) is (2 -> 1 -> 3). The number of edges on this path is 2, thus, that is the answer.
Example2

Input: A = [1, 2, 1 ,2, 2] and E = [1, 2, 1, 3, 2, 4, 2, 5]
Output: 2
Explanation:
described tree appears as follows:
1 （value = 1） / \ (value = 2) 2 3 (value = 1) / \ (value = 2) 4 5 (value = 2)
The longest path (in which all nodes have the save value of 2) is (4 -> 2 -> 5). The number of edges on this path is 2, thus, that is the answer.
Notice
Assume that: 1 <= N <= 1000

解法1：DFS
注意dfs的写法必须是dfs(…, len + 1)，而不能是
len++;
dfs(…, len)。
代码如下:

class Solution {
public:
    /**
     * @param A: as indicated in the description
     * @param E: as indicated in the description
     * @return: Return the number of edges on the longest path with same value.
     */
    int LongestPathWithSameValue(vector<int> &A, vector<int> &E) {
        int Asize = A.size();
        if (Asize == 0) return 0;
        int Esize = E.size();
        if (Esize != (Asize - 1) * 2) return 0;
        
        maxLength = 0;
        visited.resize(Asize + 1, 0);
        
        for (int i = 0; i < Esize; i += 2) {
            mp[E[i]].insert(E[i + 1]);
            mp[E[i + 1]].insert(E[i]);
        }

        for (int i = 1; i <= Asize; ++i) {
            dfs(A, E, i, 0);
        }
        
        return maxLength;
    }

private:
    void dfs(vector<int> &A, vector<int> &E, int startNode, int  len) {
        if (visited[startNode]) return;
        visited[startNode] = 1;
        
        maxLength = max(maxLength, len);
        
        for (auto node : mp[startNode]) {
            if (A[node - 1] == A[startNode - 1]) {
                dfs(A, E, node, len + 1);
            }
        }
        visited[startNode] = 0;
    }
    
    map<int, set<int>> mp;
    int maxLength;
    vector<int> visited;
};