diff --git a/apps/mtm/models.py b/apps/mtm/models.py
index 939ddab4..a2de8d52 100644
--- a/apps/mtm/models.py
+++ b/apps/mtm/models.py
@@ -294,9 +294,10 @@ class Route(CommonADModel):
             # 3. 检查final_material_out是否是终点
             final_id = final_material_out.id
             if final_id in reverse_graph:
-                raise ParseError(
-                    f"最终物料 {final_material_out.name}(ID:{final_id}) 不能作为任何Route的输入"
-                )
+                # raise ParseError(
+                #     f"最终物料 {final_material_out.name}(ID:{final_id}) 不能作为任何Route的输入"
+                # )
+                raise ParseError("最终物料不可作为输入")
 
             # 4. BFS检查路径可达性（使用material_id操作）
             visited = set()
@@ -313,10 +314,11 @@ class Route(CommonADModel):
             # 5. 检查未到达的物料
             unreachable_ids = all_material_ids - visited
             if unreachable_ids:
-                unreachable_materials = Material.objects.filter(id__in=unreachable_ids).values_list('name', flat=True)
-                raise ParseError(
-                    f"以下物料无法到达最终物料: {list(unreachable_materials)}"
-                )
+                # unreachable_materials = Material.objects.filter(id__in=unreachable_ids).values_list('name', flat=True)
+                # raise ParseError(
+                #     f"以下物料无法到达最终物料: {list(unreachable_materials)}"
+                # )
+                raise ParseError("存在无法到达的节点")
 
             # 6. DFS检查循环依赖
             visited_cycle = set()
@@ -337,8 +339,9 @@ class Route(CommonADModel):
 
             for material_id in all_material_ids:
                 if has_cycle(material_id):
-                    cycle_material = Material.objects.get(id=material_id)
-                    raise ParseError(f"循环依赖涉及物料: {cycle_material.name}(ID:{material_id})")
+                    # cycle_material = Material.objects.get(id=material_id)
+                    # raise ParseError(f"循环依赖涉及物料: {cycle_material.name}(ID:{material_id})")
+                    raise ParseError('存在循环依赖')
 
             return True