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Description |
题目描述 |
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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） |
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Input |
输入 |
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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。 |
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Output |
输出 |
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For each test case, output the maximum value. |
每个测试样例输出其最大价值。 |
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Sample Input - 输入样例 |
Sample Output - 输出样例 |
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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++】
1 #include<cstdio> 2 #include<cstring> 3 int main(){ 4 int weight[5001], t, i, j, n, B, max_value, w, v; 5 scanf("%d", &t); 6 7 while (t--){ 8 scanf("%d%d", &n, &B); 9 memset(weight, 0, sizeof(weight)); 10 weight[0] = B, max_value = 0; 11 12 for (j = 0; j < n; ++j){ 13 scanf("%d%d", &w, &v); 14 for (i = max_value; i >= 0; --i){ 15 if (weight[i] - w > weight[i + v]) weight[i + v] = weight[i] - w; 16 } 17 for (i = max_value + 1; i <= 5000; ++i) if (weight[i]) max_value = i; 18 } 19 20 printf("%d ", max_value); 21 } 22 return 0; 23 }
FZU 2214