FZU 2214 Knapsack problem（背包问题）

Description	题目描述
Given a set of n items, each with a weight w[i] and a value v[i], determine a way to choose the items into a knapsack so that the total weight is less than or equal to a given limit B and the total value is as large as possible. Find the maximum total value. (Note that each item can be only chosen once).	给你n件物品，以及每件物品的质量w[i]和价值v[i]。选择一种装包方式使得背包的最终质量小等于上限B并且最终价值尽可能大。找出最大的总价值。（注意，每件物品只能被选择一次）

Input

输入

The first line contains the integer T indicating to the number of test cases. For each test case, the first line contains the integers n and B. Following n lines provide the information of each item. The i-th line contains the weight w[i] and the value v[i] of the i-th item respectively. 1 <= number of test cases <= 100 1 <= n <= 500 1 <= B, w[i] <= 1000000000 1 <= v[1]+v[2]+...+v[n] <= 5000 All the inputs are integers.

输入的首行是一个整数T表示测试样例的数量。 对于每个测试样例，第一行包含两个整数n和B。 接下来有n行表示每件物品的信息。 第i行分别包含第i件物品的质量w[i]与价值v[i]。 1 <= 测试样例数量 <= 100 1 <= n <= 500 1 <= B, w[i] <= 1000000000 1 <= v[1]+v[2]+...+v[n] <= 5000 输入均为整数。

Output	输出
For each test case, output the maximum value.	每个测试样例输出其最大价值。

Sample Input - 输入样例	Sample Output - 输出样例
1 5 15 12 4 2 2 1 1 4 10 1 2	15

【题解】

最大质量为1000000000，数组肯定不够用。

不过，总价值才5000，我们以价值为轴开辟记录剩余可载质量的一维数组，后面的做法就与01背包如出一辙。

【代码 C++】

 #include<cstdio>
 #include<cstring>
 int main(){
     int weight[], t, i, j, n, B, max_value, w, v;
     scanf("%d", &t);

     while (t--){
         scanf("%d%d", &n, &B);
         memset(weight, , sizeof(weight));
         weight[] = B, max_value = ;

         for (j = ; j < n; ++j){
             scanf("%d%d", &w, &v);
             for (i = max_value; i >= ; --i){
                 if (weight[i] - w > weight[i + v]) weight[i + v] = weight[i] - w;
             }
             for (i = max_value + ; i <= ; ++i) if (weight[i]) max_value = i;
         }

         printf("%d\n", max_value);
     }
     return ;
 }

FZU 2214