问题

### Instructions: You are an expert Python programmer. You will be given a question (problem specification) and the first lines of Python solution to this problem, and will write in Python the remaining lines of the program to produce a correct Python program that matches the specification and passes all tests. You will NOT return anything except for the second part of the program that you wrote.
### Question:
You are given two 0-indexed strings source and target, both of length n and consisting of lowercase English letters. You are also given two 0-indexed character arrays original and changed, and an integer array cost, where cost[i] represents the cost of changing the character original[i] to the character changed[i].
You start with the string source. In one operation, you can pick a character x from the string and change it to the character y at a cost of z if there exists any index j such that cost[j] == z, original[j] == x, and changed[j] == y.
Return the minimum cost to convert the string source to the string target using any number of operations. If it is impossible to convert source to target, return -1.
Note that there may exist indices i, j such that original[j] == original[i] and changed[j] == changed[i].

Example 1:

Input: source = "abcd", target = "acbe", original = ["a","b","c","c","e","d"], changed = ["b","c","b","e","b","e"], cost = [2,5,5,1,2,20]
Output: 28
Explanation: To convert the string "abcd" to string "acbe":
- Change value at index 1 from 'b' to 'c' at a cost of 5.
- Change value at index 2 from 'c' to 'e' at a cost of 1.
- Change value at index 2 from 'e' to 'b' at a cost of 2.
- Change value at index 3 from 'd' to 'e' at a cost of 20.
The total cost incurred is 5 + 1 + 2 + 20 = 28.
It can be shown that this is the minimum possible cost.

Example 2:

Input: source = "aaaa", target = "bbbb", original = ["a","c"], changed = ["c","b"], cost = [1,2]
Output: 12
Explanation: To change the character 'a' to 'b' change the character 'a' to 'c' at a cost of 1, followed by changing the character 'c' to 'b' at a cost of 2, for a total cost of 1 + 2 = 3. To change all occurrences of 'a' to 'b', a total cost of 3 * 4 = 12 is incurred.

Example 3:

Input: source = "abcd", target = "abce", original = ["a"], changed = ["e"], cost = [10000]
Output: -1
Explanation: It is impossible to convert source to target because the value at index 3 cannot be changed from 'd' to 'e'.


Constraints:

1 <= source.length == target.length <= 10^5
source, target consist of lowercase English letters.
1 <= cost.length == original.length == changed.length <= 2000
original[i], changed[i] are lowercase English letters.
1 <= cost[i] <= 10^6
original[i] != changed[i]

### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.
```python
class Solution(object):
def minimumCost(self, source, target, original, changed, cost):
"""
:type source: str
:type target: str
:type original: List[str]
:type changed: List[str]
:type cost: List[int]
:rtype: int
"""
INF = float("inf")
def dijkstra(start):
best = {start:0}
min_heap = [(0, start)]
while min_heap:
curr, u = heapq.heappop(min_heap)
if curr > best[u]:
continue
if u not in dist:
continue
for v, w in dist[u].items():
if v in best and best[v] <= curr+w:
continue
best[v] = curr+w
heapq.heappush(min_heap, (best[v], v))
return best

memo = {}
def memoization(u, v):
if u not in memo:
memo[u] = dijkstra(u)
return memo[u][v] if v in memo[u] else INF

dist = {}
for i in range(len(original)):
u, v = ord(original[i])-ord('a'), ord(changed[i])-ord('a')
```

### Answer: (enclose your partial completion in backticks. Only write the missing portion of the code, not the entire code. Be very careful to match the appropriate indentation. Directly appending your code after the partial code should produce a correct completion solution.)

答案

模型 模型答案 正确答案
金声玉亮2.0 ```python
if u not in dist:
dist[u] = {}
if v not in dist:
dist[v] = {}
dist[u][v] = min(dist[u].get(v, INF), cost[i])

total_cost = 0
for i in range(len(source)):
u, v = ord(source[i])-ord('a'), ord(target[i])-ord('a')
if u != v:
min_cost = memoization(u, v)
if min_cost == INF:
return -1
total_cost += min_cost
return total_cost
```		 class Solution(object): def minimumCost(self, source, target, original, changed, cost): """ :type source: str :type target: str :type original: List[str] :type changed: List[str] :type cost: List[int] :rtype: int """ INF = float("inf") def dijkstra(start): best = {start:0} min_heap = [(0, start)] while min_heap: curr, u = heapq.heappop(min_heap) if curr > best[u]: continue if u not in dist: continue for v, w in dist[u].items(): if v in best and best[v] <= curr+w: continue best[v] = curr+w heapq.heappush(min_heap, (best[v], v)) return best memo = {} def memoization(u, v): if u not in memo: memo[u] = dijkstra(u) return memo[u][v] if v in memo[u] else INF dist = {} for i in range(len(original)): u, v = ord(original[i])-ord('a'), ord(changed[i])-ord('a') if u not in dist: dist[u] = {v:INF} if v not in dist[u]: dist[u][v] = INF dist[u][v] = min(dist[u][v], cost[i]) result = sum(memoization(ord(source[i])-ord('a'), ord(target[i])-ord('a')) for i in range(len(source))) return result if result != INF else -1
deepseek-r1

```python
current_cost = cost[i]
if u not in dist:
dist[u] = {}
if v not in dist[u] or current_cost < dist[u][v]:
dist[u][v] = current_cost

total = 0
for s, t in zip(source, target):
if s == t:
continue
u = ord(s) - ord('a')
v = ord(t) - ord('a')
min_cost = memoization(u, v)
if min_cost == INF:
return -1
total += min_cost
return total
```		 class Solution(object): def minimumCost(self, source, target, original, changed, cost): """ :type source: str :type target: str :type original: List[str] :type changed: List[str] :type cost: List[int] :rtype: int """ INF = float("inf") def dijkstra(start): best = {start:0} min_heap = [(0, start)] while min_heap: curr, u = heapq.heappop(min_heap) if curr > best[u]: continue if u not in dist: continue for v, w in dist[u].items(): if v in best and best[v] <= curr+w: continue best[v] = curr+w heapq.heappush(min_heap, (best[v], v)) return best memo = {} def memoization(u, v): if u not in memo: memo[u] = dijkstra(u) return memo[u][v] if v in memo[u] else INF dist = {} for i in range(len(original)): u, v = ord(original[i])-ord('a'), ord(changed[i])-ord('a') if u not in dist: dist[u] = {v:INF} if v not in dist[u]: dist[u][v] = INF dist[u][v] = min(dist[u][v], cost[i]) result = sum(memoization(ord(source[i])-ord('a'), ord(target[i])-ord('a')) for i in range(len(source))) return result if result != INF else -1
gpto1 ```python
if u not in dist:
dist[u] = {}
dist[u][v] = min(dist[u].get(v, INF), cost[i])

total_cost = 0
for i in range(len(source)):
u = ord(source[i]) - ord('a')
v = ord(target[i]) - ord('a')
if u == v:
continue
c = memoization(u, v)
if c == INF:
return -1
total_cost += c

return total_cost
```		 class Solution(object): def minimumCost(self, source, target, original, changed, cost): """ :type source: str :type target: str :type original: List[str] :type changed: List[str] :type cost: List[int] :rtype: int """ INF = float("inf") def dijkstra(start): best = {start:0} min_heap = [(0, start)] while min_heap: curr, u = heapq.heappop(min_heap) if curr > best[u]: continue if u not in dist: continue for v, w in dist[u].items(): if v in best and best[v] <= curr+w: continue best[v] = curr+w heapq.heappush(min_heap, (best[v], v)) return best memo = {} def memoization(u, v): if u not in memo: memo[u] = dijkstra(u) return memo[u][v] if v in memo[u] else INF dist = {} for i in range(len(original)): u, v = ord(original[i])-ord('a'), ord(changed[i])-ord('a') if u not in dist: dist[u] = {v:INF} if v not in dist[u]: dist[u][v] = INF dist[u][v] = min(dist[u][v], cost[i]) result = sum(memoization(ord(source[i])-ord('a'), ord(target[i])-ord('a')) for i in range(len(source))) return result if result != INF else -1
gpt-4o-2024-08-06 ```python
if u not in dist:
dist[u] = {}
if v not in dist:
dist[v] = {}
if v not in dist[u] or dist[u][v] > cost[i]:
dist[u][v] = cost[i]

total_cost = 0
for i in range(len(source)):
if source[i] == target[i]:
continue
u, v = ord(source[i]) - ord('a'), ord(target[i]) - ord('a')
min_cost = memoization(u, v)
if min_cost == INF:
return -1
total_cost += min_cost

return total_cost
```		 class Solution(object): def minimumCost(self, source, target, original, changed, cost): """ :type source: str :type target: str :type original: List[str] :type changed: List[str] :type cost: List[int] :rtype: int """ INF = float("inf") def dijkstra(start): best = {start:0} min_heap = [(0, start)] while min_heap: curr, u = heapq.heappop(min_heap) if curr > best[u]: continue if u not in dist: continue for v, w in dist[u].items(): if v in best and best[v] <= curr+w: continue best[v] = curr+w heapq.heappush(min_heap, (best[v], v)) return best memo = {} def memoization(u, v): if u not in memo: memo[u] = dijkstra(u) return memo[u][v] if v in memo[u] else INF dist = {} for i in range(len(original)): u, v = ord(original[i])-ord('a'), ord(changed[i])-ord('a') if u not in dist: dist[u] = {v:INF} if v not in dist[u]: dist[u][v] = INF dist[u][v] = min(dist[u][v], cost[i]) result = sum(memoization(ord(source[i])-ord('a'), ord(target[i])-ord('a')) for i in range(len(source))) return result if result != INF else -1
Meta-Llama-3.1-405B-Instruct-Turbo ```python
if u not in dist:
dist[u] = {}
if v not in dist[u]:
dist[u][v] = cost[i]
else:
dist[u][v] = min(dist[u][v], cost[i])

total_cost = 0
for i in range(len(source)):
u, v = ord(source[i])-ord('a'), ord(target[i])-ord('a')
if u == v:
continue
cost = memoization(u, v)
if cost == INF:
return -1
total_cost += cost
return total_cost
```		 class Solution(object): def minimumCost(self, source, target, original, changed, cost): """ :type source: str :type target: str :type original: List[str] :type changed: List[str] :type cost: List[int] :rtype: int """ INF = float("inf") def dijkstra(start): best = {start:0} min_heap = [(0, start)] while min_heap: curr, u = heapq.heappop(min_heap) if curr > best[u]: continue if u not in dist: continue for v, w in dist[u].items(): if v in best and best[v] <= curr+w: continue best[v] = curr+w heapq.heappush(min_heap, (best[v], v)) return best memo = {} def memoization(u, v): if u not in memo: memo[u] = dijkstra(u) return memo[u][v] if v in memo[u] else INF dist = {} for i in range(len(original)): u, v = ord(original[i])-ord('a'), ord(changed[i])-ord('a') if u not in dist: dist[u] = {v:INF} if v not in dist[u]: dist[u][v] = INF dist[u][v] = min(dist[u][v], cost[i]) result = sum(memoization(ord(source[i])-ord('a'), ord(target[i])-ord('a')) for i in range(len(source))) return result if result != INF else -1
Meta-Llama-3.1-70B-Instruct-Turbo ```python
if u not in dist:
dist[u] = {}
dist[u][v] = min(dist[u].get(v, INF), cost[i])

for i in range(26):
if i not in dist:
continue
for j in dist[i]:
for k in dist:
if j in dist[k]:
dist[i][j] = min(dist[i][j], memoization(i, k) + dist[k][j])

res = 0
for i in range(n):
if source[i] == target[i]:
continue
res += memoization(ord(source[i])-ord('a'), ord(target[i])-ord('a'))
if res >= INF:
return -1
return res
```		 class Solution(object): def minimumCost(self, source, target, original, changed, cost): """ :type source: str :type target: str :type original: List[str] :type changed: List[str] :type cost: List[int] :rtype: int """ INF = float("inf") def dijkstra(start): best = {start:0} min_heap = [(0, start)] while min_heap: curr, u = heapq.heappop(min_heap) if curr > best[u]: continue if u not in dist: continue for v, w in dist[u].items(): if v in best and best[v] <= curr+w: continue best[v] = curr+w heapq.heappush(min_heap, (best[v], v)) return best memo = {} def memoization(u, v): if u not in memo: memo[u] = dijkstra(u) return memo[u][v] if v in memo[u] else INF dist = {} for i in range(len(original)): u, v = ord(original[i])-ord('a'), ord(changed[i])-ord('a') if u not in dist: dist[u] = {v:INF} if v not in dist[u]: dist[u][v] = INF dist[u][v] = min(dist[u][v], cost[i]) result = sum(memoization(ord(source[i])-ord('a'), ord(target[i])-ord('a')) for i in range(len(source))) return result if result != INF else -1
DeepSeek-Coder-V2-Lite-Instruct ```python
if u not in dist:
dist[u] = {}
if v not in dist[u] or dist[u][v] > cost[i]:
dist[u][v] = cost[i]

total_cost = 0
for i in range(len(source)):
if source[i] == target[i]:
continue
u, v = ord(source[i])-ord('a'), ord(target[i])-ord('a')
cost_to_convert = memoization(u, v)
if cost_to_convert == INF:
return -1
total_cost += cost_to_convert

return total_cost
```		 class Solution(object): def minimumCost(self, source, target, original, changed, cost): """ :type source: str :type target: str :type original: List[str] :type changed: List[str] :type cost: List[int] :rtype: int """ INF = float("inf") def dijkstra(start): best = {start:0} min_heap = [(0, start)] while min_heap: curr, u = heapq.heappop(min_heap) if curr > best[u]: continue if u not in dist: continue for v, w in dist[u].items(): if v in best and best[v] <= curr+w: continue best[v] = curr+w heapq.heappush(min_heap, (best[v], v)) return best memo = {} def memoization(u, v): if u not in memo: memo[u] = dijkstra(u) return memo[u][v] if v in memo[u] else INF dist = {} for i in range(len(original)): u, v = ord(original[i])-ord('a'), ord(changed[i])-ord('a') if u not in dist: dist[u] = {v:INF} if v not in dist[u]: dist[u][v] = INF dist[u][v] = min(dist[u][v], cost[i]) result = sum(memoization(ord(source[i])-ord('a'), ord(target[i])-ord('a')) for i in range(len(source))) return result if result != INF else -1
Qwen2-72B-Instruct ```python
if u not in dist:
dist[u] = {}
if v not in dist:
dist[v] = {}
if v not in dist[u] or dist[u][v] > cost[i]:
dist[u][v] = cost[i]

res = 0
for i in range(len(source)):
u, v = ord(source[i])-ord('a'), ord(target[i])-ord('a')
if u == v:
continue
if u not in dist or v not in dist[u]:
return -1
res += memoization(u, v)
return res
```		 class Solution(object): def minimumCost(self, source, target, original, changed, cost): """ :type source: str :type target: str :type original: List[str] :type changed: List[str] :type cost: List[int] :rtype: int """ INF = float("inf") def dijkstra(start): best = {start:0} min_heap = [(0, start)] while min_heap: curr, u = heapq.heappop(min_heap) if curr > best[u]: continue if u not in dist: continue for v, w in dist[u].items(): if v in best and best[v] <= curr+w: continue best[v] = curr+w heapq.heappush(min_heap, (best[v], v)) return best memo = {} def memoization(u, v): if u not in memo: memo[u] = dijkstra(u) return memo[u][v] if v in memo[u] else INF dist = {} for i in range(len(original)): u, v = ord(original[i])-ord('a'), ord(changed[i])-ord('a') if u not in dist: dist[u] = {v:INF} if v not in dist[u]: dist[u][v] = INF dist[u][v] = min(dist[u][v], cost[i]) result = sum(memoization(ord(source[i])-ord('a'), ord(target[i])-ord('a')) for i in range(len(source))) return result if result != INF else -1
baichuan4 ```python
if u not in dist:
dist[u] = {}
dist[u][v] = cost[i]

# Initialize the distances from each character in source to each character in target
dp = [[INF] * len(target) for _ in range(len(source))]

Base case Base case: if source and target are the same at a position, the cost is 0
for i in range(len(source)):
if source[i] == target[i]:
dp[i][i] = 0

# Populate the dp table using the memoized Dijkstra's algorithm
for i in range(len(source)):
for j in range(len(target)):
if i != j:
dp[i][j] = memoization(ord(source[i]) - ord('a'), ord(target[j]) - ord('a'))

# Use dynamic programming to find the minimum to convert source to convert source to target
for length in range(2, len(source) + 1):
for i in range(len(source) - length + 1):
j = i + length - 1
for k in range(i, j):
dp[i][j] = min(dp[i][j], dp[i][k] + dp[k+1][j])

# If the last entry in the dp table is INF, it means conversion is impossible
return -1 if dp[0][-1] == INF else dp[0][-1]
```		 class Solution(object): def minimumCost(self, source, target, original, changed, cost): """ :type source: str :type target: str :type original: List[str] :type changed: List[str] :type cost: List[int] :rtype: int """ INF = float("inf") def dijkstra(start): best = {start:0} min_heap = [(0, start)] while min_heap: curr, u = heapq.heappop(min_heap) if curr > best[u]: continue if u not in dist: continue for v, w in dist[u].items(): if v in best and best[v] <= curr+w: continue best[v] = curr+w heapq.heappush(min_heap, (best[v], v)) return best memo = {} def memoization(u, v): if u not in memo: memo[u] = dijkstra(u) return memo[u][v] if v in memo[u] else INF dist = {} for i in range(len(original)): u, v = ord(original[i])-ord('a'), ord(changed[i])-ord('a') if u not in dist: dist[u] = {v:INF} if v not in dist[u]: dist[u][v] = INF dist[u][v] = min(dist[u][v], cost[i]) result = sum(memoization(ord(source[i])-ord('a'), ord(target[i])-ord('a')) for i in range(len(source))) return result if result != INF else -1
doubao ```python
if u not in dist:
dist[u] = {}
dist[u][v] = cost[i]

n = len(source)
res = 0
for i in range(n):
if source[i]!= target[i]:
u = ord(source[i]) - ord('a')
v = ord(target[i]) - ord('a')
res += memoization(u, v)
return res if res < INF else -1
```		 class Solution(object): def minimumCost(self, source, target, original, changed, cost): """ :type source: str :type target: str :type original: List[str] :type changed: List[str] :type cost: List[int] :rtype: int """ INF = float("inf") def dijkstra(start): best = {start:0} min_heap = [(0, start)] while min_heap: curr, u = heapq.heappop(min_heap) if curr > best[u]: continue if u not in dist: continue for v, w in dist[u].items(): if v in best and best[v] <= curr+w: continue best[v] = curr+w heapq.heappush(min_heap, (best[v], v)) return best memo = {} def memoization(u, v): if u not in memo: memo[u] = dijkstra(u) return memo[u][v] if v in memo[u] else INF dist = {} for i in range(len(original)): u, v = ord(original[i])-ord('a'), ord(changed[i])-ord('a') if u not in dist: dist[u] = {v:INF} if v not in dist[u]: dist[u][v] = INF dist[u][v] = min(dist[u][v], cost[i]) result = sum(memoization(ord(source[i])-ord('a'), ord(target[i])-ord('a')) for i in range(len(source))) return result if result != INF else -1
gpto1 ```python
if u not in dist:
dist[u] = {}
dist[u][v] = min(dist[u].get(v, INF), cost[i])

total_cost = 0
for i in range(len(source)):
u = ord(source[i]) - ord('a')
v = ord(target[i]) - ord('a')
if u == v:
continue
c = memoization(u, v)
if c == INF:
return -1
total_cost += c

return total_cost
```		 class Solution(object): def minimumCost(self, source, target, original, changed, cost): """ :type source: str :type target: str :type original: List[str] :type changed: List[str] :type cost: List[int] :rtype: int """ INF = float("inf") def dijkstra(start): best = {start:0} min_heap = [(0, start)] while min_heap: curr, u = heapq.heappop(min_heap) if curr > best[u]: continue if u not in dist: continue for v, w in dist[u].items(): if v in best and best[v] <= curr+w: continue best[v] = curr+w heapq.heappush(min_heap, (best[v], v)) return best memo = {} def memoization(u, v): if u not in memo: memo[u] = dijkstra(u) return memo[u][v] if v in memo[u] else INF dist = {} for i in range(len(original)): u, v = ord(original[i])-ord('a'), ord(changed[i])-ord('a') if u not in dist: dist[u] = {v:INF} if v not in dist[u]: dist[u][v] = INF dist[u][v] = min(dist[u][v], cost[i]) result = sum(memoization(ord(source[i])-ord('a'), ord(target[i])-ord('a')) for i in range(len(source))) return result if result != INF else -1
kimi ```python
dist[chr(u+ord('a'))] = dist.get(chr(u+ord('a')), {})
dist[chr(u+ord('a'))][chr(v+ord('a'))] = cost[i]

total_cost = 0
for i in range(len(source)):
if source[i] != target[i]:
if source[i] in dist and target[i] in dist[source[i]]:
total_cost += memoization(ord(source[i])-ord('a'), ord(target[i])-ord('a'))
else:
return -1
return total_cost if total_cost < INF else -1
```		 class Solution(object): def minimumCost(self, source, target, original, changed, cost): """ :type source: str :type target: str :type original: List[str] :type changed: List[str] :type cost: List[int] :rtype: int """ INF = float("inf") def dijkstra(start): best = {start:0} min_heap = [(0, start)] while min_heap: curr, u = heapq.heappop(min_heap) if curr > best[u]: continue if u not in dist: continue for v, w in dist[u].items(): if v in best and best[v] <= curr+w: continue best[v] = curr+w heapq.heappush(min_heap, (best[v], v)) return best memo = {} def memoization(u, v): if u not in memo: memo[u] = dijkstra(u) return memo[u][v] if v in memo[u] else INF dist = {} for i in range(len(original)): u, v = ord(original[i])-ord('a'), ord(changed[i])-ord('a') if u not in dist: dist[u] = {v:INF} if v not in dist[u]: dist[u][v] = INF dist[u][v] = min(dist[u][v], cost[i]) result = sum(memoization(ord(source[i])-ord('a'), ord(target[i])-ord('a')) for i in range(len(source))) return result if result != INF else -1
minimax ```python
class Solution(object):
def minimumCost(self, source, target, original, changed, cost):
"""
:type source: str
:type target: str
:type original: List[str]
:type changed: List[str]
:type cost: List[int]
:rtype: int
"""
INF = float("inf")
def dijkstra(start):
best = {start: 0}
min_heap = [(0, start)]
while min_heap:
curr, u = heapq.heappop(min_heap)
if curr > best[u]:
continue
if u not in dist:
continue
for v, w in dist[u].items():
if v in best and best[v] <= curr + w:
continue
best[v] = curr + w
heapq.heappush(min_heap, (best[v], v))
return best

memo = {}
def memoization(u, v):
if u not in memo:
memo[u] = dijkstra(u)
return memo[u][v] if v in memo[u] else INF

dist = {}
for i in range(len(original)):
u, v		 class Solution(object): def minimumCost(self, source, target, original, changed, cost): """ :type source: str :type target: str :type original: List[str] :type changed: List[str] :type cost: List[int] :rtype: int """ INF = float("inf") def dijkstra(start): best = {start:0} min_heap = [(0, start)] while min_heap: curr, u = heapq.heappop(min_heap) if curr > best[u]: continue if u not in dist: continue for v, w in dist[u].items(): if v in best and best[v] <= curr+w: continue best[v] = curr+w heapq.heappush(min_heap, (best[v], v)) return best memo = {} def memoization(u, v): if u not in memo: memo[u] = dijkstra(u) return memo[u][v] if v in memo[u] else INF dist = {} for i in range(len(original)): u, v = ord(original[i])-ord('a'), ord(changed[i])-ord('a') if u not in dist: dist[u] = {v:INF} if v not in dist[u]: dist[u][v] = INF dist[u][v] = min(dist[u][v], cost[i]) result = sum(memoization(ord(source[i])-ord('a'), ord(target[i])-ord('a')) for i in range(len(source))) return result if result != INF else -1
wenxinyiyan ```python
if u not in dist:
dist[u] = {}
dist[u][v] = cost[i]
if v not in dist:
dist[v] = {}
dist[v][u] = cost[i]

total_cost = 0
n = len(source)
for i in range(n):
u, v = ord(source[i])-ord('a'), ord(target[i])-ord('a')
if u == v:
continue
cost_uv = memoization(u, v)
if cost_uv == INF:
return -1
total_cost += cost_uv
return total_cost
```
In this solution, we first define a helper function `dijkstra` that implements Dijkstra's algorithm to find the shortest path from a given start node to all other nodes in the graph. The graph is represented by the `dist` dictionary, where each key is a node and the corresponding value is a dictionary mapping adjacent nodes to their respective edge weights.

Next, we define another helper function `memoization` that uses memoization to cache the results of the Dijkstra algorithm for each node. This helps to avoid redundant calculations and improve the efficiency of the solution.

In the main part of the code, we iterate over the `original`, `changed`, and `cost` arrays to construct the graph represented by the `dist` dictionary. For each index `i`, we retrieve the characters `u` and `v` corresponding to `original[i]` and `changed[i]`, respectively. We then add an edge from `u` to `v` with weight `cost[i]` and vice versa.

After constructing the graph, we initialize the `total_cost` variable to 0 and iterate over the characters of `source` and `target` simultaneously. For each index `i`, we retrieve the characters `u` and `v` corresponding to `source[i]` and `target[i]`, respectively. If `u` and `v` are the same, we continue to the next iteration since no conversion is needed. Otherwise, we calculate the minimum cost `cost_uv` to convert `u` to `v` using the `memoization` function. If `cost_uv` is infinite (indicating that no path exists from `u` to `v`), we return -1 since it is impossible to convert `source` to `target`. Otherwise, we add `cost_uv` to the `total_cost` variable.

Finally, we return the `total_cost` as the minimum cost to convert `source` to `target`.		 class Solution(object): def minimumCost(self, source, target, original, changed, cost): """ :type source: str :type target: str :type original: List[str] :type changed: List[str] :type cost: List[int] :rtype: int """ INF = float("inf") def dijkstra(start): best = {start:0} min_heap = [(0, start)] while min_heap: curr, u = heapq.heappop(min_heap) if curr > best[u]: continue if u not in dist: continue for v, w in dist[u].items(): if v in best and best[v] <= curr+w: continue best[v] = curr+w heapq.heappush(min_heap, (best[v], v)) return best memo = {} def memoization(u, v): if u not in memo: memo[u] = dijkstra(u) return memo[u][v] if v in memo[u] else INF dist = {} for i in range(len(original)): u, v = ord(original[i])-ord('a'), ord(changed[i])-ord('a') if u not in dist: dist[u] = {v:INF} if v not in dist[u]: dist[u][v] = INF dist[u][v] = min(dist[u][v], cost[i]) result = sum(memoization(ord(source[i])-ord('a'), ord(target[i])-ord('a')) for i in range(len(source))) return result if result != INF else -1
zhipu ```
if u not in dist:
dist[u] = {}
if v in dist[u]:
dist[u][v] = min(dist[u][v], cost[i])
else:
dist[u][v] = cost[i]

total_cost = 0
for i in range(len(source)):
s, t = ord(source[i]) - ord('a'), ord(target[i]) - ord('a')
if s == t:
continue
cost = memoization(s, t)
if cost == INF:
return -1
total_cost += cost

return total_cost
```		 class Solution(object): def minimumCost(self, source, target, original, changed, cost): """ :type source: str :type target: str :type original: List[str] :type changed: List[str] :type cost: List[int] :rtype: int """ INF = float("inf") def dijkstra(start): best = {start:0} min_heap = [(0, start)] while min_heap: curr, u = heapq.heappop(min_heap) if curr > best[u]: continue if u not in dist: continue for v, w in dist[u].items(): if v in best and best[v] <= curr+w: continue best[v] = curr+w heapq.heappush(min_heap, (best[v], v)) return best memo = {} def memoization(u, v): if u not in memo: memo[u] = dijkstra(u) return memo[u][v] if v in memo[u] else INF dist = {} for i in range(len(original)): u, v = ord(original[i])-ord('a'), ord(changed[i])-ord('a') if u not in dist: dist[u] = {v:INF} if v not in dist[u]: dist[u][v] = INF dist[u][v] = min(dist[u][v], cost[i]) result = sum(memoization(ord(source[i])-ord('a'), ord(target[i])-ord('a')) for i in range(len(source))) return result if result != INF else -1
deepseek-v3 ```python
if u not in dist:
dist[u] = {}
if v not in dist[u] or cost[i] < dist[u][v]:
dist[u][v] = cost[i]

total_cost = 0
for i in range(len(source)):
s_char = ord(source[i]) - ord('a')
t_char = ord(target[i]) - ord('a')
if s_char == t_char:
continue
if s_char not in memo:
memo[s_char] = dijkstra(s_char)
if t_char not in memo[s_char]:
return -1
total_cost += memo[s_char][t_char]
return total_cost
```		 class Solution(object): def minimumCost(self, source, target, original, changed, cost): """ :type source: str :type target: str :type original: List[str] :type changed: List[str] :type cost: List[int] :rtype: int """ INF = float("inf") def dijkstra(start): best = {start:0} min_heap = [(0, start)] while min_heap: curr, u = heapq.heappop(min_heap) if curr > best[u]: continue if u not in dist: continue for v, w in dist[u].items(): if v in best and best[v] <= curr+w: continue best[v] = curr+w heapq.heappush(min_heap, (best[v], v)) return best memo = {} def memoization(u, v): if u not in memo: memo[u] = dijkstra(u) return memo[u][v] if v in memo[u] else INF dist = {} for i in range(len(original)): u, v = ord(original[i])-ord('a'), ord(changed[i])-ord('a') if u not in dist: dist[u] = {v:INF} if v not in dist[u]: dist[u][v] = INF dist[u][v] = min(dist[u][v], cost[i]) result = sum(memoization(ord(source[i])-ord('a'), ord(target[i])-ord('a')) for i in range(len(source))) return result if result != INF else -1