spring如何实现两个xml配置文件间的互调

package soundsystem;
public class Dog {
  private String Cry;
  private Cat Cat;
  public void setCry(String cry) {
    Cry = cry;
  }
  public void setCat(soundsystem.Cat cat) {
    Cat = cat;
  }
  public void DogCry(){
    System.out.println("狗叫："+Cry);
    Cat.CatCry();
  }
}

package soundsystem;
public class Cat {
  private String Cry;
  public Cat(String cry){
    this.Cry=cry;
  }
  public void CatCry(){
    System.out.println("猫叫："+Cry);
  }
}

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd">

  <import resource="Bean_CatXML.xml"></import>

  <bean id="Dog" class="soundsystem.Dog">
    <property name="Cry" value="汪汪汪~"></property>
    <property name="cat" ref="Cat"></property>
  </bean>

</beans>

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd">

  <bean id="Cat" class="soundsystem.Cat">
    <constructor-arg value="喵喵~"></constructor-arg>
  </bean>
</beans>

package Test;

import org.junit.runner.RunWith;
import org.springframework.context.ApplicationContext;
import org.springframework.context.support.ClassPathXmlApplicationContext;
import org.springframework.test.context.junit4.SpringJUnit4ClassRunner;
import soundsystem.Cat;
import soundsystem.Dog;
@RunWith(SpringJUnit4ClassRunner.class)
public class Test {
  @org.junit.Test
  public static void main(String[] args) {
    ApplicationContext ap=new ClassPathXmlApplicationContext("config/Bean_DogXML.xml");
    Dog dog=(Dog)ap.getBean("Dog");
    dog.DogCry();
  }
}