Find the minimum spanning tree with alternating colored edges
Given a graph with N nodes and M edges where each edge has a color (either black or green) and a cost associated with it. Find the minimum spanning tree of the graph such that every path in the tree is made up of alternating colored edges.
Examples:
Input: N = 3, M = 4
Output: 6Input: N = 4, M = 6
Output: 4
Approach:
- The first observation we make here is every such kind of spanning tree will be a chain. To prove it, suppose we have a tree that is not a chain and every path in it is made up of alternating edges. Then we can deduce that at least 1 node has a degree of 3. Out of these 3 edges, at least 2 will have the same color. The path using these 2 edges will never follow the conditions and Hence, such kind of tree is always a chain.
- Now we can find a chain with minimum cost and alternating edges using bitmask-dp,
dp[mask(2^n)][Node(n)][col_of_last_edge(2)] where the mask is the bitmask of nodes we’ve added to the chain. Node is the last node we added to the chain.col_of_last_edge is the color of edge use to connect Node. - To transition from 1 state to another state we visit the adjacency list of the last node and use those edges which have color != col_of_last_edge.
Below is the implementation of the above approach:
C++
// C++ program for the // above approach #include <bits/stdc++.h> using namespace std; int graph[18][18][2]; // Initializing dp of size = // (2^18)*18*2. long long dp[1 << 18][18][2]; // Recursive Function to calculate // Minimum Cost with alternate // colour edges long long minCost(int n, int m, int mask, int prev, int col) { // Base case if (mask == ((1 << n) - 1)) { return 0; } // If already calculated if (dp[mask][prev][col == 1] != 0) { return dp[mask][prev][col == 1]; } long long ans = 1e9; for (int i = 0; i < n; i++) { for (int j = 0; j < 2; j++) { // Masking previous edges // as explained in above formula. if (graph[prev][i][j] && !(mask & (1 << i)) && (j != col)) { long long z = graph[prev][i][j] + minCost(n,m,mask|(1<<i),i,j); ans = min(z, ans); } } } return dp[mask][prev][col == 1] = ans; } // Function to Adjacency // List Representation // of a Graph void makeGraph(vector<pair<pair<int,int>, pair<int,char>>>& vp,int m){ for (int i = 0; i < m; i++) { int a = vp[i].first.first - 1; int b = vp[i].first.second - 1; int cost = vp[i].second.first; char color = vp[i].second.second; graph[a][b][color == 'W'] = cost; graph[b][a][color == 'W'] = cost; } } // Function to getCost // for the Minimum Spanning // Tree Formed int getCost(int n,int m){ // Assigning maximum // possible value. long long ans = 1e9; for (int i = 0; i < n; i++) { ans = min(ans, minCost(n, m, 1 << i, i, 2)); } if (ans != 1e9) { return ans; } else { return -1; } } // Driver code int main() { int n = 3, m = 4; vector<pair<pair<int, int>, pair<int, char> > > vp = { { { 1, 2 }, { 2, 'B' } }, { { 1, 2 }, { 3, 'W' } }, { { 2, 3 }, { 4, 'W' } }, { { 2, 3 }, { 5, 'B' } } }; makeGraph(vp,m); cout << getCost(n,m) << '\n'; return 0; } |
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Python3
# Python implementation of the approach graph = [[[0, 0] for i in range(18)] for j in range(18)] # Initializing dp of size = # (2^18)*18*2. dp = [[[0, 0] for i in range(18)] for j in range(1 << 15)] # Recursive Function to calculate # Minimum Cost with alternate # colour edges def minCost(n: int, m: int, mask: int, prev: int, col: int) -> int: global dp # Base case if mask == ((1 << n) - 1): return 0 # If already calculated if dp[mask][prev][col == 1] != 0: return dp[mask][prev][col == 1] ans = int(1e9) for i in range(n): for j in range(2): # Masking previous edges # as explained in above formula. if graph[prev][i][j] and not (mask & (1 << i)) \ and (j != col): z = graph[prev][i][j] + minCost(n, m, mask | (1 << i), i, j) ans = min(z, ans) dp[mask][prev][col == 1] = ans return dp[mask][prev][col == 1] # Function to Adjacency # List Representation # of a Graph def makeGraph(vp: list, m: int): global graph for i in range(m): a = vp[i][0][0] - 1 b = vp[i][0][1] - 1 cost = vp[i][1][0] color = vp[i][1][1] graph[a][b][color == 'W'] = cost graph[b][a][color == 'W'] = cost # Function to getCost # for the Minimum Spanning # Tree Formed def getCost(n: int, m: int) -> int: # Assigning maximum # possible value. ans = int(1e9) for i in range(n): ans = min(ans, minCost(n, m, 1 << i, i, 2)) if ans != int(1e9): return ans else: return -1 # Driver Code if __name__ == "__main__": n = 3 m = 4 vp = [[[1, 2], [2, 'B']], [[1, 2], [3, 'W']], [[2, 3], [4, 'W']], [[2, 3], [5, 'B']]] makeGraph(vp, m) print(getCost(n, m)) # This code is contributed by # sanjeev2552 |
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Output:
6
Time Complexity: O(2^N * (M + N))
