java - GOTO/继续 Java 动态编程

Question

我有以下 Java 代码实现动态编程递归关系:

public double routeCost() throws Exception {
    double cost = Double.MAX_VALUE;
    for (int l=i; l<=j; l++) {
        if (! (customers.get(l) instanceof VehicleCustomer) )
            continue;
        double value = F(l,j) + (customers.get(l).distanceFrom(depot));
        if (value < cost)
            cost = value;
    }

    return cost;
}

private double F(int l, int m) {

    //=========================== FIRST CASE ===========================
    if (l==i && m==i) {
        //System.out.println(i+","+j+","+l+","+m);
        return firstCase();
    }

    //=========================== SECOND CASE ===========================
    if (l==i && (i<m && m<=j) ) {
        //System.out.println(i+","+j+","+l+","+m);
        //analyses the possibility of performing all the soubtours based at heicle customert_i
        return secondCase(i,m);

    }
    //=========================== GENERAL CASE ===========================
    else  {
        System.out.println(i+","+j+","+l+","+m);

        assert (customers.get(l) instanceof VehicleCustomer);

        assert ( (i<l && l<=j) && (l<=m && m<=j) );
        return Math.min(thirdCaseFirstTerm(l,m), thirdCaseSecondTerm(l,m));
    } 

}

private double firstCase() {
    mainRoute.add(depot);
    mainRoute.add(customers.get(i));
    return depot.distanceFrom(customers.get(i));
}

private double secondCase(int i,int m) {
    double caseValue = Double.MAX_VALUE;
    int k = i;
    while (k<m) {   
        double totalDemand=0;
        for (int u=k+1; ( (u<=m) && (totalDemand<=truckCapacity) ); u++)
            totalDemand += customers.get(u).getDemand();

        double cost = F(i,k) + thita(i,k+1,m);
        if (cost <= caseValue) 
            caseValue = cost;

        k++;
    }       
    return caseValue;
}

private double thirdCaseFirstTerm(int l, int m) {
    double caseValue = Double.MAX_VALUE;

    int k = i;
    while (k<m) {

        double totalDemand=0;
        for (int u=k+1; ( (u<=m) && (totalDemand<=truckCapacity) ); u++)
            totalDemand += customers.get(u).getDemand();

        double cost = F(l,k) + thita(l,k+1,m);
        if (cost <= caseValue) 
            caseValue = cost;
        k++;
    }

    return caseValue;
}

private double thirdCaseSecondTerm(int l,int m) {
    double caseValue = Double.MAX_VALUE;

    int k = i; 

    for (Customer cust : customers) {
        int h = customers.indexOf(cust);
        if ( (!(cust instanceof VehicleCustomer)) || (h >=l)) {
            continue;
        }

        double totalDemand=0;
        for (int u=k+2; ( (u<=m) && (totalDemand<=truckCapacity) ); u++)
            totalDemand += customers.get(u).getDemand();

        double cost = F(h,k) + customers.get(h).distanceFrom(customers.get(l)) + thita(l,k+2,m);
        if (cost < caseValue)
            caseValue = cost;
    }

    return caseValue;
}

方法F(int,int)是从方法routeCost()的for循环中调用的。我想找到一种方法来强制执行断言 assert (customers.get(l) instanceof VehicleCustomer);` 不是 true，我不想继续执行 return 语句，而是想从 RouteCost() 进入 for 循环以继续下一次迭代。但 F() 必须返回一个值!

我知道我想做的事情几乎违反了面向对象的所有规则，但我真的需要它。

Answer 1

您可以在 F() 中引发 Exception 并在 routeCost() 中捕获它。

这种方法比使用断言要好得多。它们在实践中很少使用，这是有充分理由的:异常更加灵活，更适合检测错误、无效输入等。

^{_{PS:当我说“很少使用”时，我的说法是基于这样一个事实:我在过去几年中看到了数十万行 Java 代码，并且很少遇到使用断言的代码.}}