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用回溯解背包问题 假设有n件物品，定义一个结构体a[]来存储，结构体有两个成员weight和value(weight表示重量，value表示价值)先定义一个数组col[]表示每个物品当前状态（为1表示被选，为0表示未被选），其初值全为1，从下标为0开始遍历，当前所选物品总重和总价值分别设为tw和tv（初值均为0），背包的限重设为limit，若第i个物品满足tw+a[i].weight<=limit且col[i]==1 就将a[i].weight和value加入tw和tv，否则col[i]设为0。-Knapsack problem using backtracking solution assumption has n items, the definition of a structure a [] to store, structure, body weight and has two members value (weight, said weight, value, said value) before the definition of an array col [] mean that each and every item current status (as one said to be elected, not selected to express to 0), all of its initial value 1, from the beginning subscript 0 ergodicity, the currently selected items and the total value of gross weight, respectively, as tw and tv (early values are 0), backpack weight limit is set to limit, if paragraph i of goods to meet tw+ a [i]. weight
