2203. Minimum Weighted Subgraph With the Required Paths
Description
You are given an integer n
denoting the number of nodes of a weighted directed graph. The nodes are numbered from 0
to n - 1
.
You are also given a 2D integer array edges
where edges[i] = [fromi, toi, weighti]
denotes that there exists a directed edge from fromi
to toi
with weight weighti
.
Lastly, you are given three distinct integers src1
, src2
, and dest
denoting three distinct nodes of the graph.
Return the minimum weight of a subgraph of the graph such that it is possible to reach dest
from both src1
and src2
via a set of edges of this subgraph. In case such a subgraph does not exist, return -1
.
A subgraph is a graph whose vertices and edges are subsets of the original graph. The weight of a subgraph is the sum of weights of its constituent edges.
Example 1:
Input: n = 6, edges = [[0,2,2],[0,5,6],[1,0,3],[1,4,5],[2,1,1],[2,3,3],[2,3,4],[3,4,2],[4,5,1]], src1 = 0, src2 = 1, dest = 5 Output: 9 Explanation: The above figure represents the input graph. The blue edges represent one of the subgraphs that yield the optimal answer. Note that the subgraph [[1,0,3],[0,5,6]] also yields the optimal answer. It is not possible to get a subgraph with less weight satisfying all the constraints.
Example 2:
Input: n = 3, edges = [[0,1,1],[2,1,1]], src1 = 0, src2 = 1, dest = 2 Output: -1 Explanation: The above figure represents the input graph. It can be seen that there does not exist any path from node 1 to node 2, hence there are no subgraphs satisfying all the constraints.
Constraints:
3 <= n <= 105
0 <= edges.length <= 105
edges[i].length == 3
0 <= fromi, toi, src1, src2, dest <= n - 1
fromi != toi
src1
,src2
, anddest
are pairwise distinct.1 <= weight[i] <= 105
Solution
class Solution:
def minimumWeight(self, n: int, edges: List[List[int]], src1: int, src2: int, dest: int) -> int:
G = defaultdict(list)
RG = defaultdict(list)
for x, y, w in edges:
G[x].append((y, w))
RG[y].append((x, w))
def dijkstra(graph, src):
dist = [float('inf')] * n
dist[src] = 0
pq = [(0, src)]
while pq:
d, node = heapq.heappop(pq)
if dist[node] != d: continue
for nei, w in graph[node]:
old = dist[nei]
new = d + w
if new < old:
dist[nei] = new
heapq.heappush(pq, (new, nei))
return dist
A = dijkstra(G, src1)
B = dijkstra(G, src2)
C = dijkstra(RG, dest)
res = float('inf')
for a, b, c in zip(A, B, C):
res = min(res, a + b + c)
return -1 if res == float('inf') else res