From 6c497c692b53da154458e82d719cc9c25268e132 Mon Sep 17 00:00:00 2001
From: swjtuhjf <www.690924@qq.com>
Date: Tue, 12 Nov 2024 14:04:38 +0800
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---
 .../0096.城市间货物运输III.md          | 45 +++++++++++++++++++
 1 file changed, 45 insertions(+)

diff --git a/problems/kamacoder/0096.城市间货物运输III.md b/problems/kamacoder/0096.城市间货物运输III.md
index a41332cf..109027b1 100644
--- a/problems/kamacoder/0096.城市间货物运输III.md
+++ b/problems/kamacoder/0096.城市间货物运输III.md
@@ -703,6 +703,9 @@ public class Main {
 ```
 
 ### Python
+
+Bellman-Ford方法求解单源有限最短路
+
 ```python
 def main():
     # 輸入
@@ -736,6 +739,48 @@ def main():
             
             
 
+if __name__ == "__main__":
+    main()
+```
+
+SPFA方法求解单源有限最短路
+
+```python
+from collections import deque
+from math import inf
+
+
+def main():
+    n, m = [int(i) for i in input().split()]
+    graph = [[] for _ in range(n+1)]
+    for _ in range(m):
+        v1, v2, val = [int(i) for i in input().split()]
+        graph[v1].append([v2, val])
+    src, dst, k = [int(i) for i in input().split()]
+    min_dist = [inf for _ in range(n+1)]
+    min_dist[src] = 0  # 初始化起点的距离
+    que = deque([src])
+
+    while k != -1 and que:
+        visited = [False for _ in range(n+1)]  # 用于保证每次松弛时一个节点最多加入队列一次
+        que_size = len(que)
+        temp_dist = min_dist.copy()  # 用于记录上一次遍历的结果
+        for _ in range(que_size):
+            cur_node = que.popleft()
+            for next_node, val in graph[cur_node]:
+                if min_dist[next_node] > temp_dist[cur_node] + val:
+                    min_dist[next_node] = temp_dist[cur_node] + val
+                    if not visited[next_node]:
+                        que.append(next_node)
+                        visited[next_node] = True
+        k -= 1
+
+    if min_dist[dst] == inf:
+        print("unreachable")
+    else:
+        print(min_dist[dst])
+
+
 if __name__ == "__main__":
     main()
 ```