两个包的背包算法

Question

我找到了thread，它为带有 2 个背包的背包算法提供了伪代码。 我已经尝试在 C++ 中实现它，但它不像假设的那样工作。代码如下：

#include <cstdio>
#define MAX_W1 501
#define MAX_W2 501

int maximum(int a, int b, int c) {
    int max = a>b?a:b;
    return c>max?c:max;
}

int knapsack[MAX_W1][MAX_W2] = {0};

int main() {
    int n, s1, s2, gain, weight; // items, sack1, sack2, gain, cost

    scanf("%d %d %d", &n, &s1, &s2);

    // filing knapsack
    for (int i = 0; i < n; i++) {
        scanf("%d %d", &gain, &weight);

        for (int w1 = s1; w1 >= weight; w1--) {
            for (int w2 = s2; w2 >= weight; w2--) {
                knapsack[w1][w2] = maximum(
                    knapsack[w1][w2],                 // we have best option
                    knapsack[w1 - weight][w2] + gain, // put into sack one
                    knapsack[w1][w2 - weight] + gain  // put into sack two
                );
            }
        }
    }

    int result = 0;

    // searching for result
    for (int i = 0; i <= s1; i++) {
        for (int j = 0; j <= s2; j++) {
            if (knapsack[i][j] > result) {
                result = knapsack[i][j];
            }
        }
    }

    printf("%d\n", result);

    return 0;
}

例如以下输入：

5 4 3
6 2
3 2
4 1
2 1
1 1

我有输出：

13

显然这是错误的，因为我可以将所有物品（1,2 放入第一个袋子，其余放入第二个袋子），总和为 16。 如果我得到伪代码错误的任何解释，我将不胜感激。

我几乎没有更新，因为有些人在理解输入格式方面有问题：

1. 第一行包含 3 个数字，分别是物品数量、一袋容量、二袋容量
2. 后来有 n 行，每行包含 2 个数字：增益，第 i 个项目的成本。
3. 假设麻袋不能大于 500。

Answer 1

您使用的算法似乎不正确，因为它只会考虑对象恰好适合两个麻袋的情况。我对您的代码进行了以下更改，它现在可以正常运行：

#include <algorithm>

using std::max;

int max3(int a, int b, int c) {
    return max(a, max(b, c));
}

和

for (int w1 = s1; w1 >= 0; w1--) {
    for (int w2 = s2; w2 >= 0; w2--) {
        if (w1 >= weight && w2 >= weight) // either sack has room
        {
            knapsack[w1][w2] = max3(
                knapsack[w1][w2],                 // we have best option
                knapsack[w1 - weight][w2] + gain, // put into sack one
                knapsack[w1][w2 - weight] + gain  // put into sack two
            );
        }
        else if (w1 >= weight) // only sack one has room
        {
            knapsack[w1][w2] = max(
                knapsack[w1][w2],                 // we have best option
                knapsack[w1 - weight][w2] + gain  // put into sack one
            );
        }
        else if (w2 >= weight) // only sack two has room
        {
            knapsack[w1][w2] = max(
                knapsack[w1][w2],                 // we have best option
                knapsack[w1][w2 - weight] + gain  // put into sack two
            );
        }
    }
}

Answer 2

这里是修改代码以使其工作：-

#include <cstdio>
#define MAX_W1 501
#define MAX_W2 501

int maximum(int a, int b, int c) {
    int max = a>b?a:b;
    return c>max?c:max;
}

int knapsack[MAX_W1][MAX_W2] = {0};

int main() {
    int n, s1, s2, gain, weight; // items, sack1, sack2, gain, cost

    scanf("%d %d %d", &n, &s1, &s2);

    // filing knapsack
    for (int i = 0; i < n; i++) {
        scanf("%d %d", &gain, &weight);
    // need to fill up all the table cannot stop if one sack is full because item might fit in other
        for (int w1 = s1; w1 >= 0; w1--) {
            for (int w2 = s2; w2 >= 0; w2--) {
                 int val1=0,val2=0;
                 if(weight<=w1)
                   val1 = knapsack[w1 - weight][w2] + gain;
                 if(weight<=w2)
                   val2 = knapsack[w1][w2 - weight] + gain;

                 knapsack[w1][w2] = maximum(
                    knapsack[w1][w2],                   // we have best option
                     val1,              // put into sack one
                     val2               // put into sack two
                  );


            }
        }
    }


    // No need to search for max value it always be Knapsack[s1][s2]
    printf("%d\n", knapsack[s1][s2]);

    return 0;
}