首先抽象一下问题：

给定集合C和阈值m，要求找出C的所有符合条件的子集S，使得S的所有元素相加之和为m.

对于这个问题，其实一样可以通过动态规划进行求解。求解过程可以采用递归和非递归的形式，而一般所谓的递归形式(特例：[尾递归]，可以通过编译器优化，转换成非递归形式)，其实就是调用栈重复直接或间接调用自身的过程。所以通过栈的数据结构完全可以解决，而且这种解决方式，在一定程度上来说，就是把递归求解过程通过栈进行展开。闲话少说，上代码(C++好久没写了，改用Java语言写了一个实现，为了直观体现求解过程，代码中忽略了一些常规的数据校验)：

package com.lee.test; import java.util.ArrayList; import java.util.Arrays; import java.util.List; public class PackageProblem { public static void main(String[] args) { int[] problemSpace = {1, 8, 4, 3, 5, 2}; int threshold = 10; printProblem(problemSpace, threshold); printSolution(solve(problemSpace, threshold)); } private static void printProblem(int[] problemSpace, int threshold) { System.out.println("Problem Space : "+Arrays.toString(problemSpace)); System.out.println("Problem threshold : "+threshold); } private static void printSolution(List solutions) { System.out.println("Solutions : "); if(solutions == null || solutions.size() < 1) { System.out.println("no solution."); }else { for(int[] solution : solutions) { System.out.println(Arrays.toString(solution)); } } } static class Element { int index; int value; Element(int index, int value) { this.index = index; this.value = value; } } static class Stack { private Element[] elements; private int top; private int sum; Stack(int capacity) { elements = new Element[capacity]; top = -1; sum = 0; } void push(int index, int value) { if(top == elements.length - 1) { throw new IllegalStateException("stack is already full."); } elements[++top] = new Element(index, value); sum += value; } Element pop() { if(top == -1) { throw new IllegalStateException("stack is already empty."); } Element element = elements[top--]; sum -= element.value; return element; } boolean isEmpty() { return top == -1; } int sum() { return sum; } int[] snapshot() { int[] snapshot = new int[top+1]; for(int i=0; i<= top; i++) { snapshot[i] = elements[i].value; } return snapshot; } } private static List solve(int[] problemSpace, int threshold) { if(problemSpace == null) { throw new NullPointerException("problem space is null."); } List solutions = new ArrayList(); if(problemSpace.length < 1) { return solutions; } Stack stack = new Stack(problemSpace.length); stack.push(0, problemSpace[0]); int index = 0; while(true) { int sum = stack.sum(); if(sum >= threshold) { if(sum == threshold) { solutions.add(stack.snapshot()); } index = stack.pop().index; } if(index < problemSpace.length - 1) { stack.push(++index, problemSpace[index]); }else { if(sum < threshold) { stack.pop(); } if(stack.isEmpty()) { break; } index = stack.pop().index; } } return solutions; } }

PS: 自己运行了一下，应该是Ok，有什么问题，欢迎留言