diff --git a/apps/pm/services.py b/apps/pm/services.py
index be120c0e..0e5dc3de 100644
--- a/apps/pm/services.py
+++ b/apps/pm/services.py
@@ -15,35 +15,51 @@ class PmService:
 
     @classmethod
     def cal_x_task_count(cls, target_materialId:str, target_quantity, route_qs):
-        # 构建逆向依赖图 {输出物料: [所有能生产它的工序]}
+        # 构建逆向依赖图 {输出物料ID: [Route]}
         graph = defaultdict(list)
         for route in route_qs:
-            graph[route.material_out.id].append(route)
+            if route.material_out:  # 确保输出物料存在
+                graph[route.material_out.id].append(route)
 
-        # 存储每个Route记录的生产数量 {route_id: 需生产数量}
+        # 存储每个物料的需求量
+        demands = defaultdict(float)
+        demands[target_materialId] = target_quantity
         step_production = defaultdict(int)
 
-        # 广度优先逆向遍历（BFS）
-        queue = deque([(target_materialId, target_quantity)])
+        # 逆向遍历，从目标物料开始
+        queue = deque([target_materialId])
         while queue:
-            current_material, current_demand = queue.popleft()
-            for route in graph.get(current_material, []):
-                # 根据工序类型计算当前工序的生产量
-                if route.process.mtype == 20:  # 拆分
-                    production = current_demand * route.div_number / (route.out_rate / 100)
-                elif route.process.mtype == 30:  # 合并
-                    production = current_demand / route.div_number / (route.out_rate / 100)
-                else:  # 正常
-                    production = current_demand / (route.out_rate / 100)
+            current_out_id = queue.popleft()
+            current_demand = demands.get(current_out_id, 0)
+            if current_demand <= 0:
+                continue
 
-                # 记录当前工序的生产量（向上取整）
-                step_production[route.id] = math.ceil(production)
+            # 处理所有能生产该物料的Route（多路径时需求均分）
+            routes = graph.get(current_out_id, [])
+            if not routes:
+                continue
 
-                # 将输入物料需求加入队列
+            # 均分需求到所有Route（示例按均分逻辑，可根据业务调整）
+            demand_per_route = math.ceil(current_demand / len(routes)) if len(routes) > 1 else current_demand
+
+            for route in routes:
+                # 记录该Route的产出量（向上取整）
+                step_production[route.id] = math.ceil(demand_per_route)
+
+                # 根据工序类型计算输入需求
+                if route.process.mtype == 20:  # 拆分工序：1个输入拆成多个输出
+                    input_needed = demand_per_route / route.div_number / (route.out_rate / 100)
+                elif route.process.mtype == 30:  # 合并工序：多个输入合并为1个输出
+                    input_needed = demand_per_route * route.div_number / (route.out_rate / 100)
+                else:  # 正常工序
+                    input_needed = demand_per_route / (route.out_rate / 100)
+
+                # 更新上游物料需求
                 if route.material_in:
-                    queue.append((route.material_in.id, production))
+                    demands[route.material_in.id] += input_needed
+                    queue.append(route.material_in.id)
 
-        return dict(step_production)
+        return dict(reversed(step_production.items()))
 
 
     @classmethod