Number of Unicolored Paths between two nodes
Given an undirected colored graph(edges are colored), with a source vertex ‘s’ and a destination vertex ‘d’, print number of paths which from given ‘s’ to ‘d’ such that the path is UniColored(edges in path having same color).
The edges are colored, here Colors are represented with numbers. At maximum, number of different colors will be number of edges.
Input : Graph
u, v, color
1, 2, 1
1, 3, 2
2, 3, 3
2, 4, 2
2, 5, 4
3, 5, 3
4, 5, 2
source = 2 destination = 5
Output : 3
Explanation : There are three paths from 2 to 5
2 -> 5 with color red
2 -> 3 - > 5 with color sky blue
2 -> 4 - > 5 with color green
Algorithm :
1. Do dfs traversal on the neighbour nodes of source node.
2. The color between source node and neighbour nodes is known, if the DFS traversal also have same color, proceed, else stop going on that path.
3. After reaching destination node, increment count by 1.
NOTE : Number of Colors will always be less than number of edges.
C++
// C++ code to find unicolored paths #include <bits/stdc++.h> using namespace std; const int MAX_V = 100; int color[MAX_V]; bool vis[MAX_V]; // Graph class represents a udirected graph // using adjacency list representation class Graph { // vertices, edges, adjancy list int V; int E; vector<pair<int, int> > adj[MAX_V]; // function used by UniColorPaths // DFS traversal o from x to y void dfs(int x, int y, int z); // Constructor public: Graph(int V, int E); // function to add an edge to graph void addEdge(int v, int w, int z); // finds paths between a and b having // same color edges int UniColorPaths(int a, int b); }; Graph::Graph(int V, int E) { this -> V = V; this -> E = E; } void Graph::addEdge(int a, int b, int c) { adj[a].push_back({b, c}); // Add b to a’s list. adj[b].push_back({a, c}); // Add c to b’s list. } void Graph::dfs(int x, int y, int col) { if (vis[x]) return; vis[x] = 1; // mark this as a possible color to reach s to d if (x == y) { color[col] = 1; return; } // if the next edge is also of same color for (int i = 0; i < int(adj[x].size()); i++) if (adj[x][i].second == col) dfs(adj[x][i].first, y, col); } // function that finds paths between a and b // such that all edges are same colored // It uses recursive dfs() int Graph::UniColorPaths(int a, int b) { // dfs on nodes directly connected to source for (int i = 0; i < int(adj[a].size()); i++) { dfs(a, b, adj[a][i].second); // to visit again visited nodes memset(vis, 0, sizeof(vis)); } int cur = 0; for (int i = 0; i <= E; i++) cur += color[i]; return (cur); } // driver code int main() { // Create a graph given in the above diagram Graph g(5, 7); g.addEdge(1, 2, 1); g.addEdge(1, 3, 2); g.addEdge(2, 3, 3); g.addEdge(2, 4, 2); g.addEdge(2, 5, 4); g.addEdge(3, 5, 3); g.addEdge(4, 5, 2); int s = 2; // source int d = 5; // destination cout << "Number of unicolored paths : "; cout << g.UniColorPaths(s, d) << endl; return 0; } |
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Python3
# Python3 code to find unicolored paths MAX_V = 100 color = [0] * MAX_V vis = [0] * MAX_V # Graph class represents a udirected graph # using adjacency list representation class Graph: def __init__(self, V, E): self.V = V self.E = E self.adj = [[] for i in range(MAX_V)] # Function used by UniColorPaths # DFS traversal o from x to y def dfs(self, x, y, col): if vis[x]: return vis[x] = 1 # mark this as a possible color to reach s to d if x == y: color[col] = 1 return # if the next edge is also of same color for i in range(0, len(self.adj[x])): if self.adj[x][i][1] == col: self.dfs(self.adj[x][i][0], y, col) def addEdge(self, a, b, c): self.adj[a].append((b, c)) # Add b to a’s list. self.adj[b].append((a, c)) # Add c to b’s list. # Function that finds paths between a # and b such that all edges are same # colored. It uses recursive dfs() def UniColorPaths(self, a, b): global vis # dfs on nodes directly connected to source for i in range(0, len(self.adj[a])): self.dfs(a, b, self.adj[a][i][1]) # to visit again visited nodes vis = [0] * len(vis) cur = 0 for i in range(0, self.E + 1): cur += color[i] return cur # Driver code if __name__ == "__main__": # Create a graph given in the above diagram g = Graph(5, 7) g.addEdge(1, 2, 1) g.addEdge(1, 3, 2) g.addEdge(2, 3, 3) g.addEdge(2, 4, 2) g.addEdge(2, 5, 4) g.addEdge(3, 5, 3) g.addEdge(4, 5, 2) s = 2 # source d = 5 # destination print("Number of unicolored paths : ", end = "") print(g.UniColorPaths(s, d)) # This code is contributed by Rituraj Jain |
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Output:
Number of unicolored paths : 3
Time Complexity : O(E * (E + V))
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