Given a tree with N nodes. The task is to color the tree with the minimum number of colors(K) such that the colors of the edges incident to a vertex are different. Print K in first-line and then in next line print N – 1 space-separated integer represents the colors of the edges.
Examples:
Input: N = 3, edges[][] = {{0, 1}, {1, 2}}
0
/
1
/
2
Output:
2
1 2
0
/ (1)
1
/ (2)
2
Input: N = 8, edges[][] = {{0, 1}, {1, 2},
{1, 3}, {1, 4},
{3, 6}, {4, 5},
{5, 7}}
0
/
1
/ \ \
2 3 4
/ \
6 5
\
7
Output:
4
1 2 3 4 1 1 2
Approach: First, let’s think about the minimum number of colors K. For every vertex v, deg(v) ≤ K should meet (where deg(v) denotes the degree of vertex v). In fact, there exists a vertex with all K different colors. First, choose a vertex and let it be the root, thus T will be a rooted tree. Perform a breadth-first search from the root. For each vertex, determine the colors of edges between its children one by one. For each edge, use the color with the minimum index among those which are not used as colors of edges whose one of endpoints is the current vertex. Then each index of color does not exceed K.
Below is the implementation of the above approach:
CPP
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Add an edge between the vertexes void add_edge(vector<vector<int> >& gr, int x, int y, vector<pair<int, int> >& edges) { gr[x].push_back(y); gr[y].push_back(x); edges.push_back({ x, y }); } // Function to color the tree with minimum // number of colors such that the colors of // the edges incident to a vertex are different int color_tree(int n, vector<vector<int> >& gr, vector<pair<int, int> >& edges) { // To store the minimum colors int K = 0; // To store color of the edges map<pair<int, int>, int> color; // Color of edge between its parent vector<int> cs(n, 0); // To check if the vertex is // visited or not vector<int> used(n, 0); queue<int> que; used[0] = 1; que.emplace(0); while (!que.empty()) { // Take first element of the queue int v = que.front(); que.pop(); // Take the possible value of K if (K < (int)gr[v].size()) K = gr[v].size(); // Current color int cur = 1; for (int u : gr[v]) { // If vertex is already visited if (used[u]) continue; // If the color is similar // to it's parent if (cur == cs[v]) cur++; // Assign the color cs[u] = color[make_pair(u, v)] = color[make_pair(v, u)] = cur++; // Mark it visited used[u] = 1; // Push into the queue que.emplace(u); } } // Print the minimum required colors cout << K << endl; // Print the edge colors for (auto p : edges) cout << color[p] << " "; } // Driver code int main() { int n = 8; vector<vector<int> > gr(n); vector<pair<int, int> > edges; // Add edges add_edge(gr, 0, 1, edges); add_edge(gr, 1, 2, edges); add_edge(gr, 1, 3, edges); add_edge(gr, 1, 4, edges); add_edge(gr, 3, 6, edges); add_edge(gr, 4, 5, edges); add_edge(gr, 5, 7, edges); // Function call color_tree(n, gr, edges); return 0; } |
Python
# Python3 implementation of the approach from collections import deque as queue gr = [[] for i in range(100)] edges = [] # Add an edge between the vertexes def add_edge(x, y): gr[x].append(y) gr[y].append(x) edges.append([x, y]) # Function to color the tree with minimum # number of colors such that the colors of # the edges incident to a vertex are different def color_tree(n): # To store the minimum colors K = 0 # To store color of the edges color = dict() # Color of edge between its parent cs = [0] * (n) # To check if the vertex is # visited or not used = [0] * (n) que = queue() used[0] = 1 que.append(0) while (len(que) > 0): # Take first element of the queue v = que.popleft() # Take the possible value of K if (K < len(gr[v])): K = len(gr[v]) # Current color cur = 1 for u in gr[v]: # If vertex is already visited if (used[u]): continue # If the color is similar # to it's parent if (cur == cs[v]): cur += 1 # Assign the color cs[u] = cur color[(u, v)] = color[(v, u)] = cur cur += 1 # Mark it visited used[u] = 1 # Push into the queue que.append(u) # Print minimum required colors print(K) # Print edge colors for p in edges: i = (p[0], p[1]) print(color[i], end = " ") # Driver code n = 8 # Add edges add_edge(0, 1) add_edge(1, 2) add_edge(1, 3) add_edge(1, 4) add_edge(3, 6) add_edge(4, 5) add_edge(5, 7) # Function call color_tree(n) # This code is contributed by mohit kumar 29 |
4 1 2 3 4 1 1 2
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
Recommended Posts:
- Minimum number of edges between two vertices of a Graph
- Minimum steps to reach any of the boundary edges of a matrix | Set-2
- Finding the path from one vertex to rest using BFS
- Minimum nodes to be colored in a Graph such that every node has a colored neighbour
- Print Nodes in Top View of Binary Tree
- BFS vs DFS for Binary Tree
- Check if a given graph is tree or not
- Diameter of n-ary tree using BFS
- Level of Each node in a Tree from source node (using BFS)
- Count the number of nodes at given level in a tree using BFS.
- Iterative approach to check for children sum property in a Binary Tree
- Burn the binary tree starting from the target node
- Duplicate subtree in Binary Tree | SET 2
- Find parent of each node in a tree for multiple queries
- Find distance of nodes from root in a tree for multiple queries
- Left-Right traversal of all the levels of Binary tree
- Construct Binary Tree from Ancestor Matrix | Top Down Approach
- Check if the given permutation is a valid BFS of a given Tree
- Difference between sums of odd level and even level nodes in an N-ary Tree
- Longest path in an undirected tree
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.
Improved By : mohit kumar 29
