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Program to Find Out the Minimum Cost Possible from Weighted Graph in Python
Suppose we have a 2D list of integers called edges which are a representation of an undirected graph. Every row in the input represents an edge [u, v, w] meaning nodes u and v are connected and the edge has the weight w. The graph consists of n nodes from 0 to n-1.
The cost of a path is defined here as the product of the number of edges and the maximum weight for any edge in the path. We have to find out the minimum cost possible from node 0 to node n-1, or we declare the answer as -1 if no such path exists.
So, if the input is like edges = [ [0, 2, 100], [1, 2, 200], [1, 3, 100], [2, 3, 300] ], then the output will be 600
To solve this, we will follow these steps −
graph := a new map
weights := a new map
max_weight := 0
N := 0
for each u, v, w in edges, do
insert v at the end of graph[u]
insert u at the end of graph[v]
weights[u, v] := w
weights[v, u] := w
N := maximum of N, u + 1, v + 1
max_weight := maximum of max_weight, w
result := infinity
while max_weight >= 0, do
d, weight := bfs(0, max_weight)
if d >= 0, then
result := minimum of result, d * weight
max_weight := weight - 1
otherwise,
terminate the loop
return result if result < infinity otherwise -1
Define a function bfs() . This will take root, weight_cap
target := N - 1
Q := a deque containing root, 0, 0
visited := [False] * N
visited[0] := True
while Q is not empty, do
v, d, current_weight := delete last element from Q
if v is same as N - 1, then
return d, current_weight
for each w in graph[v], do
if visited[w] is non-zero, then
continue the next iteration
new_weight := weights[v, w]
if new_weight <= weight_cap, then
visited[w] := True
add (w, d + 1, maximum of (current_weight, new_weight)) at the left of Q
return -1, -1
Example
Let us see the following implementation to get better understanding −
from collections import defaultdict, deque
class Solution:
def solve(self, edges):
graph = defaultdict(list)
weights = {}
max_weight = 0
N = 0
for u, v, w in edges:
graph[u].append(v)
graph[v].append(u)
weights[u, v] = w
weights[v, u] = w
N = max(N, u + 1, v + 1)
max_weight = max(max_weight, w)
def bfs(root, weight_cap):
target = N - 1
Q = deque([(root, 0, 0)])
visited = [False] * N
visited[0] = True
while Q:
v, d, current_weight = Q.pop()
if v == N - 1:
return d, current_weight
for w in graph[v]:
if visited[w]:
continue
new_weight = weights[v, w]
if new_weight <= weight_cap:
visited[w] = True
zQ.appendleft((w, d + 1, max(current_weight, new_weight)))
return -1, -1
result = float("inf")
while max_weight >= 0:
d, weight = bfs(0, max_weight)
if d >= 0:
result = min(result, d * weight)
max_weight = weight - 1
else:
break
return result if result < float("inf") else -1
ob = Solution()
print (ob.solve( [
[0, 2, 100],
[1, 2, 200],
[1, 3, 100],
[2, 3, 300]
]))Input
[ [0, 2, 100], [1, 2, 200], [1, 3, 100], [2, 3, 300] ]
Output
600
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