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Update
This commit is contained in:
programmercarl
@@ -7,6 +7,8 @@
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# 动态规划：关于多重背包，你该了解这些！
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本题力扣上没有原题，大家可以去[卡码网第56题](https://kamacoder.com/problempage.php?pid=1066)去练习，题意是一样的。
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之前我们已经系统的讲解了01背包和完全背包，如果没有看过的录友，建议先把如下三篇文章仔细阅读一波。
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* [动态规划：关于01背包问题，你该了解这些！](https://programmercarl.com/背包理论基础01背包-1.html)
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@@ -53,79 +55,96 @@
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毫无区别，这就转成了一个01背包问题了，且每个物品只用一次。
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这种方式来实现多重背包的代码如下：
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练习题目：[卡码网第56题，多重背包](https://kamacoder.com/problempage.php?pid=1066)
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代码如下：
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```CPP
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void test_multi_pack() {
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vector<int> weight = {1, 3, 4};
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vector<int> value = {15, 20, 30};
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vector<int> nums = {2, 3, 2};
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int bagWeight = 10;
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for (int i = 0; i < nums.size(); i++) {
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while (nums[i] > 1) { // nums[i]保留到1，把其他物品都展开
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// 超时了
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#include<iostream>
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#include<vector>
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using namespace std;
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int main() {
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int bagWeight,n;
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cin >> bagWeight >> n;
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vector<int> weight(n, 0);
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vector<int> value(n, 0);
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vector<int> nums(n, 0);
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for (int i = 0; i < n; i++) cin >> weight[i];
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for (int i = 0; i < n; i++) cin >> value[i];
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for (int i = 0; i < n; i++) cin >> nums[i];
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for (int i = 0; i < n; i++) {
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while (nums[i] > 1) { // 物品数量不是一的，都展开
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weight.push_back(weight[i]);
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value.push_back(value[i]);
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nums[i]--;
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}
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}
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vector<int> dp(bagWeight + 1, 0);
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for(int i = 0; i < weight.size(); i++) { // 遍历物品
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for(int i = 0; i < weight.size(); i++) { // 遍历物品，注意此时的物品数量不是n
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for(int j = bagWeight; j >= weight[i]; j--) { // 遍历背包容量
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dp[j] = max(dp[j], dp[j - weight[i]] + value[i]);
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}
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for (int j = 0; j <= bagWeight; j++) {
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cout << dp[j] << " ";
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}
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cout << endl;
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}
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cout << dp[bagWeight] << endl;
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}
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int main() {
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test_multi_pack();
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}
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```
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* 时间复杂度：O(m × n × k)，m：物品种类个数，n背包容量，k单类物品数量
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大家去提交之后，发现这个解法超时了，为什么呢，哪里耗时呢？
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也有另一种实现方式，就是把每种商品遍历的个数放在01背包里面在遍历一遍。
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耗时就在 这段代码：
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```CPP
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for (int i = 0; i < n; i++) {
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while (nums[i] > 1) { // 物品数量不是一的，都展开
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weight.push_back(weight[i]);
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value.push_back(value[i]);
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nums[i]--;
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}
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}
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```
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如果物品数量很多的话，C++中，这种操作十分费时，主要消耗在vector的动态底层扩容上。（其实这里也可以优化，先把 所有物品数量都计算好，一起申请vector的空间。
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这里也有另一种实现方式，就是把每种商品遍历的个数放在01背包里面在遍历一遍。
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代码如下：（详看注释）
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```CPP
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#include<iostream>
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#include<vector>
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using namespace std;
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int main() {
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int bagWeight,n;
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cin >> bagWeight >> n;
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vector<int> weight(n, 0);
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vector<int> value(n, 0);
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vector<int> nums(n, 0);
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for (int i = 0; i < n; i++) cin >> weight[i];
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for (int i = 0; i < n; i++) cin >> value[i];
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for (int i = 0; i < n; i++) cin >> nums[i];
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```CPP
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void test_multi_pack() {
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vector<int> weight = {1, 3, 4};
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vector<int> value = {15, 20, 30};
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vector<int> nums = {2, 3, 2};
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int bagWeight = 10;
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vector<int> dp(bagWeight + 1, 0);
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for(int i = 0; i < weight.size(); i++) { // 遍历物品
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for(int i = 0; i < n; i++) { // 遍历物品
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for(int j = bagWeight; j >= weight[i]; j--) { // 遍历背包容量
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// 以上为01背包，然后加一个遍历个数
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for (int k = 1; k <= nums[i] && (j - k * weight[i]) >= 0; k++) { // 遍历个数
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dp[j] = max(dp[j], dp[j - k * weight[i]] + k * value[i]);
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}
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}
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// 打印一下dp数组
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for (int j = 0; j <= bagWeight; j++) {
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cout << dp[j] << " ";
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}
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cout << endl;
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}
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cout << dp[bagWeight] << endl;
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}
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int main() {
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test_multi_pack();
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}
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```
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* 时间复杂度：O(m × n × k)，m：物品种类个数，n背包容量，k单类物品数量
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时间复杂度：O(m × n × k)，m：物品种类个数，n背包容量，k单类物品数量
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从代码里可以看出是01背包里面在加一个for循环遍历一个每种商品的数量。 和01背包还是如出一辙的。
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@@ -146,320 +165,14 @@ int main() {
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### Java：
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```Java
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public void testMultiPack1(){
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// 版本一：改变物品数量为01背包格式
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List<Integer> weight = new ArrayList<>(Arrays.asList(1, 3, 4));
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List<Integer> value = new ArrayList<>(Arrays.asList(15, 20, 30));
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List<Integer> nums = new ArrayList<>(Arrays.asList(2, 3, 2));
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int bagWeight = 10;
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for (int i = 0; i < nums.size(); i++) {
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while (nums.get(i) > 1) { // 把物品展开为i
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weight.add(weight.get(i));
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value.add(value.get(i));
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nums.set(i, nums.get(i) - 1);
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}
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}
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int[] dp = new int[bagWeight + 1];
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for(int i = 0; i < weight.size(); i++) { // 遍历物品
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for(int j = bagWeight; j >= weight.get(i); j--) { // 遍历背包容量
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dp[j] = Math.max(dp[j], dp[j - weight.get(i)] + value.get(i));
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}
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System.out.println(Arrays.toString(dp));
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}
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}
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public void testMultiPack2(){
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// 版本二：改变遍历个数
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int[] weight = new int[] {1, 3, 4};
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int[] value = new int[] {15, 20, 30};
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int[] nums = new int[] {2, 3, 2};
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int bagWeight = 10;
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int[] dp = new int[bagWeight + 1];
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for(int i = 0; i < weight.length; i++) { // 遍历物品
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for(int j = bagWeight; j >= weight[i]; j--) { // 遍历背包容量
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// 以上为01背包，然后加一个遍历个数
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for (int k = 1; k <= nums[i] && (j - k * weight[i]) >= 0; k++) { // 遍历个数
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dp[j] = Math.max(dp[j], dp[j - k * weight[i]] + k * value[i]);
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}
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System.out.println(Arrays.toString(dp));
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}
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}
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}
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```
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### Python：
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改变物品数量为01背包格式（无参版）
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```python
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def test_multi_pack():
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weight = [1, 3, 4]
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value = [15, 20, 30]
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nums = [2, 3, 2]
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bagWeight = 10
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# 将数量大于1的物品展开
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for i in range(len(nums)):
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while nums[i] > 1:
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weight.append(weight[i])
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value.append(value[i])
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nums[i] -= 1
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dp = [0] * (bagWeight + 1)
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for i in range(len(weight)): # 遍历物品
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for j in range(bagWeight, weight[i] - 1, -1): # 遍历背包容量
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dp[j] = max(dp[j], dp[j - weight[i]] + value[i])
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for j in range(bagWeight + 1):
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print(dp[j], end=" ")
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print()
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print(dp[bagWeight])
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test_multi_pack()
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```
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改变遍历个数（无参版）
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```python
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def test_multi_pack():
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weight = [1, 3, 4]
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value = [15, 20, 30]
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nums = [2, 3, 2]
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bagWeight = 10
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dp = [0] * (bagWeight + 1)
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for i in range(len(weight)): # 遍历物品
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for j in range(bagWeight, weight[i] - 1, -1): # 遍历背包容量
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# 以上为01背包，然后加一个遍历个数
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for k in range(1, nums[i] + 1): # 遍历个数
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if j - k * weight[i] >= 0:
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dp[j] = max(dp[j], dp[j - k * weight[i]] + k * value[i])
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# 打印一下dp数组
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for j in range(bagWeight + 1):
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print(dp[j], end=" ")
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print()
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print(dp[bagWeight])
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test_multi_pack()
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```
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改变物品数量为01背包格式（有参版）
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```python
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def test_multi_pack(weight, value, nums, bagWeight):
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# 将数量大于1的物品展开
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for i in range(len(nums)):
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while nums[i] > 1:
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weight.append(weight[i])
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value.append(value[i])
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nums[i] -= 1
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dp = [0] * (bagWeight + 1)
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for i in range(len(weight)): # 遍历物品
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for j in range(bagWeight, weight[i] - 1, -1): # 遍历背包容量
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dp[j] = max(dp[j], dp[j - weight[i]] + value[i])
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for j in range(bagWeight + 1):
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print(dp[j], end=" ")
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print()
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print(dp[bagWeight])
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if __name__ == "__main__":
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weight = [1, 3, 4]
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value = [15, 20, 30]
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nums = [2, 3, 2]
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bagWeight = 10
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test_multi_pack(weight, value, nums, bagWeight)
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```
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改变遍历个数（有参版）
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```python
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def test_multi_pack(weight, value, nums, bagWeight):
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dp = [0] * (bagWeight + 1)
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for i in range(len(weight)): # 遍历物品
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for j in range(bagWeight, weight[i] - 1, -1): # 遍历背包容量
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# 以上为01背包，然后加一个遍历个数
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for k in range(1, nums[i] + 1): # 遍历个数
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if j - k * weight[i] >= 0:
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dp[j] = max(dp[j], dp[j - k * weight[i]] + k * value[i])
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# 使用 join 函数打印 dp 数组
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print(' '.join(str(dp[j]) for j in range(bagWeight + 1)))
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print(dp[bagWeight])
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if __name__ == "__main__":
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weight = [1, 3, 4]
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value = [15, 20, 30]
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nums = [2, 3, 2]
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bagWeight = 10
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test_multi_pack(weight, value, nums, bagWeight)
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```
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### Go：
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```go
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package theory
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import "log"
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// 多重背包可以化解为 01 背包
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func multiplePack(weight, value, nums []int, bagWeight int) int {
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for i := 0; i < len(nums); i++ {
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for nums[i] > 1 {
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weight = append(weight, weight[i])
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value = append(value, value[i])
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nums[i]--
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}
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}
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log.Println(weight)
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log.Println(value)
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res := make([]int, bagWeight+1)
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for i := 0; i < len(weight); i++ {
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for j := bagWeight; j >= weight[i]; j-- {
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res[j] = getMax(res[j], res[j-weight[i]]+value[i])
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}
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log.Println(res)
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}
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return res[bagWeight]
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}
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```
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> 单元测试
|
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|
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```go
|
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package theory
|
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|
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import "testing"
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|
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func Test_multiplePack(t *testing.T) {
|
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type args struct {
|
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weight []int
|
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value []int
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nums []int
|
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bagWeight int
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}
|
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tests := []struct {
|
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name string
|
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args args
|
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want int
|
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}{
|
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{
|
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name: "one",
|
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args: args{
|
||||
weight: []int{1, 3, 4},
|
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value: []int{15, 20, 30},
|
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nums: []int{2, 3, 2},
|
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bagWeight: 10,
|
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},
|
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want: 90,
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},
|
||||
}
|
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for _, tt := range tests {
|
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t.Run(tt.name, func(t *testing.T) {
|
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if got := multiplePack(tt.args.weight, tt.args.value, tt.args.nums, tt.args.bagWeight); got != tt.want {
|
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t.Errorf("multiplePack() = %v, want %v", got, tt.want)
|
||||
}
|
||||
})
|
||||
}
|
||||
}
|
||||
```
|
||||
|
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> 输出
|
||||
|
||||
```
|
||||
=== RUN Test_multiplePack
|
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=== RUN Test_multiplePack/one
|
||||
2022/03/02 21:09:05 [1 3 4 1 3 3 4]
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2022/03/02 21:09:05 [15 20 30 15 20 20 30]
|
||||
2022/03/02 21:09:05 [0 15 15 15 15 15 15 15 15 15 15]
|
||||
2022/03/02 21:09:05 [0 15 15 20 35 35 35 35 35 35 35]
|
||||
2022/03/02 21:09:05 [0 15 15 20 35 45 45 50 65 65 65]
|
||||
2022/03/02 21:09:05 [0 15 30 30 35 50 60 60 65 80 80]
|
||||
2022/03/02 21:09:05 [0 15 30 30 35 50 60 60 70 80 80]
|
||||
2022/03/02 21:09:05 [0 15 30 30 35 50 60 60 70 80 80]
|
||||
2022/03/02 21:09:05 [0 15 30 30 35 50 60 60 70 80 90]
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--- PASS: Test_multiplePack (0.00s)
|
||||
--- PASS: Test_multiplePack/one (0.00s)
|
||||
PASS
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||||
```
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||||
|
||||
### TypeScript：
|
||||
|
||||
> 版本一（改变数据源）：
|
||||
|
||||
```typescript
|
||||
function testMultiPack() {
|
||||
const bagSize: number = 10;
|
||||
const weightArr: number[] = [1, 3, 4],
|
||||
valueArr: number[] = [15, 20, 30],
|
||||
amountArr: number[] = [2, 3, 2];
|
||||
for (let i = 0, length = amountArr.length; i < length; i++) {
|
||||
while (amountArr[i] > 1) {
|
||||
weightArr.push(weightArr[i]);
|
||||
valueArr.push(valueArr[i]);
|
||||
amountArr[i]--;
|
||||
}
|
||||
}
|
||||
const goodsNum: number = weightArr.length;
|
||||
const dp: number[] = new Array(bagSize + 1).fill(0);
|
||||
// 遍历物品
|
||||
for (let i = 0; i < goodsNum; i++) {
|
||||
// 遍历背包容量
|
||||
for (let j = bagSize; j >= weightArr[i]; j--) {
|
||||
dp[j] = Math.max(dp[j], dp[j - weightArr[i]] + valueArr[i]);
|
||||
}
|
||||
}
|
||||
console.log(dp);
|
||||
}
|
||||
testMultiPack();
|
||||
```
|
||||
|
||||
> 版本二（改变遍历方式）：
|
||||
|
||||
```typescript
|
||||
function testMultiPack() {
|
||||
const bagSize: number = 10;
|
||||
const weightArr: number[] = [1, 3, 4],
|
||||
valueArr: number[] = [15, 20, 30],
|
||||
amountArr: number[] = [2, 3, 2];
|
||||
const goodsNum: number = weightArr.length;
|
||||
const dp: number[] = new Array(bagSize + 1).fill(0);
|
||||
// 遍历物品
|
||||
for (let i = 0; i < goodsNum; i++) {
|
||||
// 遍历物品个数
|
||||
for (let j = 0; j < amountArr[i]; j++) {
|
||||
// 遍历背包容量
|
||||
for (let k = bagSize; k >= weightArr[i]; k--) {
|
||||
dp[k] = Math.max(dp[k], dp[k - weightArr[i]] + valueArr[i]);
|
||||
}
|
||||
}
|
||||
}
|
||||
console.log(dp);
|
||||
}
|
||||
testMultiPack();
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
Reference in New Issue
Block a user