本文章用 AcWing 题目举例
一、01背包
设 d p i , j dp_{i,j} dpi,j 代表 i 件物品 j 容量背包可装的最大价值。
则其可以由两种结果递推得到
- 第i件物品不装到背包里,前i-1件物品用了j容量的包。 d p [ i + 1 ] [ j ] = d p [ i ] [ j ] dp[i+1][j] = dp[i][j] dp[i+1][j]=dp[i][j]
- 第i件物品装到背包里,前i-1件物品用了 j − w i j-w_{i} j−wi容量的包。 d p [ i + 1 ] [ j ] = d p [ i ] [ j − v [ i + 1 ] ] + w [ i + 1 ] dp[i+1][j] =dp[i][j-v[i+1]] + w[i+1] dp[i+1][j]=dp[i][j−v[i+1]]+w[i+1]
取两种递推结果的最大值即可。
1.1 朴素二维数组
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while (scanner.hasNextInt()) {
int n = scanner.nextInt();
int total = scanner.nextInt();
int[][] dp = new int[n+1][total + 1];
for (int i = 0; i < n; i++) {
int v = scanner.nextInt();
int w = scanner.nextInt();
for (int j = 0; j <= total; j++) {
if (j < v) {
dp[i+1][j] = dp[i][j];
} else {
dp[i+1][j] = Math.max(dp[i][j-v] + w,dp[i][j]);
}
}
}
System.out.println(dp[n][total]);
}
}
}
1.2 一维优化
二维转一维的方法是画出二维表格图,看在一维滚动的情况下,如何使用旧数据。
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while (scanner.hasNextInt()) {
int n = scanner.nextInt();
int total = scanner.nextInt();
int[] dp = new int[total + 1];
for (int i = 0; i < n; i++) {
int v = scanner.nextInt();
int w = scanner.nextInt();
for (int j = total; j >= v; j--) {
dp[j] = Math.max(dp[j-v] + w,dp[j]);
}
}
System.out.println(dp[total]);
}
}
}
二、完全背包
第i件物品可以选0次、1次、2次……,递推时判断所有选法的最大值
2.1 暴力推导
可以直观地得到公式:
d p [ i + 1 ] [ j ] = M a t h . m a x ( d p [ i ] [ j − v ∗ k ] + w ∗ k , d p [ i + 1 ] [ j ] ) dp[i+1][j] = Math.max(dp[i][j-v*k] + w* k,dp[i+1][j]) dp[i+1][j]=Math.max(dp[i][j−v∗k]+w∗k,dp[i+1][j])
时间复杂度为 O ( n v s ) O(nvs) O(nvs)
import java.util.Scanner;
public class Main {
/**
* dp[i][j] = dp[i-1][j],dp[i-1][j-v[i]]+w[i],...
*/
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while (scanner.hasNextInt()) {
int n = scanner.nextInt();
int total = scanner.nextInt();
int[][] dp = new int[n+1][total+1];
for (int i = 0; i < n; i++) {
int v = scanner.nextInt();
int w = scanner.nextInt();
for (int j = 0; j <= total; j++) {
for (int k = 0; k * v <= j; k++) {
dp[i+1][j] = Math.max(dp[i][j-v*k] + w* k,dp[i+1][j]);
}
}
}
System.out.println(dp[n][total]);
}
}
}
2.2 时间优化
第i件物品取0次、1次、2次……
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dp[i+1][j] = Math.max(dp[i][j-v[i+1]*k] + w[i+1]* k), k=0,1,2,3...
dp[i+1][j]=Math.max(dp[i][j−v[i+1]∗k]+w[i+1]∗k),k=0,1,2,3...
可以转化为:第i件物品取0次和至少取1次
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dp[i+1][j] = Math.max( dp[i][j],dp[i+1][j-v[i+1]] + w[i+1])
dp[i+1][j]=Math.max(dp[i][j],dp[i+1][j−v[i+1]]+w[i+1])
import java.util.Scanner;
/**
* 完全背包
*/
public class KnapsackFull {
/**
* dp[i][j] = dp[i-1][j],dp[i][j]+w[i],...
*/
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while (scanner.hasNextInt()) {
int n = scanner.nextInt();
int total = scanner.nextInt();
int[][] dp = new int[n+1][total+1];
for (int i = 0; i < n; i++) {
int v = scanner.nextInt();
int w = scanner.nextInt();
for (int j = 0; j <= total; j++) {
if (j >= v) {
dp[i+1][j] = Math.max(dp[i+1][j-v] + w,dp[i][j]);
} else {
dp[i+1][j] = dp[i][j];
}
}
}
System.out.println(dp[n][total]);
}
}
}
2.3 一维优化
/**
* 完全背包
*/
public class KnapsackFull {
/**
* dp[i][j] = dp[i-1][j],dp[i][j]+w[i],...
*/
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while (scanner.hasNextInt()) {
int n = scanner.nextInt();
int total = scanner.nextInt();
int[] dp = new int[total+1];
for (int i = 0; i < n; i++) {
int v = scanner.nextInt();
int w = scanner.nextInt();
for (int j = 0; j <= total; j++) {
if (j >= v) {
dp[j] = Math.max(dp[j-v] + w,dp[j]);
} else {
dp[j] = dp[j];
}
}
}
System.out.println(dp[total]);
}
}
}
三、多重背包
多重背包是有限制的完全背包。
3.1 暴力求解
可以在2.1节上稍加修改,增加限制
import java.util.Scanner;
/**
* 多重背包
*/
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while (scanner.hasNextInt()) {
int n = scanner.nextInt();
int total = scanner.nextInt();
int[][] dp = new int[n+1][total+1];
for (int i = 0; i < n; i++) {
int v = scanner.nextInt();
int w = scanner.nextInt();
int s = scanner.nextInt();
for (int j = 0; j <= total; j++) {
for (int k = 0; k <= s && k * v <= j; k++) {
dp[i+1][j] = Math.max(dp[i][j-v*k] + w* k,dp[i+1][j]);
}
}
}
System.out.println(dp[n][total]);
}
}
}
3.2 二进制优化
3.3 单调队列优化法
参考 https://blog.youkuaiyun.com/weixin_43693379/article/details/89432283