如何求解从源点u出发恰好经过k条边的最短路径（不限终点）？

Hey there! Let's adjust your existing code to solve the problem of finding the shortest path starting from node u that uses exactly k edges—regardless of which node it ends at.

Understanding the Original Code #

Your current shortestPath function calculates the shortest path from u to a specific v using exactly k edges. To adapt this, we just need to compute this value for all possible end nodes and then take the minimum of those results.

Modified Implementation #

Here's how you can tweak the code:

1. Keep the core recursive function (we'll reuse it to calculate paths to each end node).
2. Add a wrapper function that iterates over all possible end nodes, computes the shortest path to each with exactly k edges, and returns the smallest value.

def shortestPath(graph, u, v, k):
    V = len(graph)
    INF = 999999999999
    # Base cases
    if k == 0 and u == v:
        return 0
    if k == 1 and graph[u][v] != INF:
        return graph[u][v]
    if k <= 0:
        return INF
    # Initialize result to infinity
    res = INF
    # Iterate through all adjacent nodes of u
    for i in range(V):
        if graph[u][i] != INF and u != i:
            rec_res = shortestPath(graph, i, v, k - 1)
            if rec_res != INF:
                res = min(res, graph[u][i] + rec_res)
    return res

def shortestPathExactlyKEdgesFromU(graph, u, k):
    V = len(graph)
    INF = 999999999999
    min_total = INF
    # Check all possible end nodes
    for v in range(V):
        path_len = shortestPath(graph, u, v, k)
        if path_len < min_total:
            min_total = path_len
    return min_total if min_total != INF else -1  # Return -1 if no such path exists

# Driver code
if __name__ == '__main__':
    INF = 999999999999
    graph = [[0, 4, 2, 6, 5],
             [INF, 0, 4, 2, 5],
             [INF, INF, 0, 4, 3],
             [INF, INF, INF, 0, 3],
             [INF, INF, INF, INF, 0]]
    u = 0
    k = 3
    result = shortestPathExactlyKEdgesFromU(graph, u, k)
    if result == -1:
        print("No valid path with exactly", k, "edges exists from node", u)
    else:
        print("Weight of the shortest path with exactly", k, "edges from node", u, "is", result)

Key Changes Explained #

• Wrapper Function: shortestPathExactlyKEdgesFromU loops through every possible end node v, uses your original function to get the path length from u to v with k edges, and tracks the minimum value.
• Dynamic V Value: Instead of hardcoding V=4, we use len(graph) to make the code work for any graph size.
• Edge Case Handling: If no valid path exists (all results are INF), we return -1 and print a clear message.

Optimization Note #

The recursive approach has a time complexity of O(V^k), which can get slow for large graphs or large k. For better performance, consider using a dynamic programming approach:

• Create a DP table where dp[m][j] represents the shortest path from u to j using exactly m edges.
• Initialize dp[0][u] = 0 (0 edges, stay at u) and all other dp[0][j] = INF.
• For each step from 1 to k, update dp[m][j] = min(dp[m-1][i] + graph[i][j]) for all nodes i.
• The answer is the minimum value in dp[k][*].

Here's a quick DP implementation example:

def shortestPathExactlyKEdgesDP(graph, u, k):
    V = len(graph)
    INF = 999999999999
    # Initialize DP table: dp[m][j] = shortest path from u to j with m edges
    dp = [[INF] * V for _ in range(k+1)]
    dp[0][u] = 0  # 0 edges, start at u
    
    for m in range(1, k+1):
        for j in range(V):
            for i in range(V):
                if dp[m-1][i] != INF and graph[i][j] != INF:
                    dp[m][j] = min(dp[m][j], dp[m-1][i] + graph[i][j])
    
    min_total = min(dp[k])
    return min_total if min_total != INF else -1

# Use it in driver code:
# result = shortestPathExactlyKEdgesDP(graph, u, k)

This DP approach runs in O(k*V²) time, which is much more efficient for larger inputs.

内容的提问来源于stack exchange，提问作者Mayank Kataria