一、問題描述
給定n種物品和一個揹包。物品i的重量是Wi,其價值爲Vi,揹包的容量爲C。應如何選擇裝入揹包的物品,使得裝入揹包中物品的總價值最大?
二、算法策略
單位價值化
三、核心代碼
static double Knapsack(double[] a, double[] b, double[] x, double C, int n) {
sort(a, b, n);
int i;
double total = 0;
for (i = 0; i < n; i++) {
if (a[i] <= C) {//如果揹包剩餘的容量大於等於第i個物體的重量
x[i] = 1; //把第i個物體整個裝進揹包
C = C - a[i]; //揹包的剩餘容量減少了第i個物體的重量
} else {
break; //退出循環
}
}
if (i < n) {//判斷是否第n個物體整個裝進去揹包裏了,如果i<=n表示否定
x[i] = C / a[i];
}
for (i = 0; i < n; i++) {
total = total + x[i] * b[i];
}
return total;
}
四、整體代碼
package 第四章;
import java.util.Scanner;
public class beibao {
public static void main(String args[]) {
int n, i, j;
double C;
System.out.println("請輸入待選擇的物品的個數:");
Scanner scanner = new Scanner(System.in);
n = scanner.nextInt();
System.out.println("請輸入揹包容量:");
C = scanner.nextDouble();
double[] a = new double[n]; //物品重量數組
double[] b = new double[n]; //物品價值數組
double[] x = new double[n];
for (i = 0; i < n; i++) {
a[i] = Math.floor(Math.random() * 20);
}
for (i = 0; i < n; i++) {
b[i] = Math.floor(Math.random() * 20);
}
System.out.println("隨機產生的物品重量爲:");
for (i = 0; i < n; i++) {
System.out.print(a[i] + " ");
}
System.out.println();
System.out.println("隨機產生的物品價值爲:");
for (i = 0; i < n; i++) {
System.out.print(b[i] + " ");
}
System.out.println();
double value = Knapsack(a, b, x, C, n);
System.out.println("最大價值爲:" + value);
}
static void sort(double[] a, double[] b, int n) {
double[] c = new double[n];
for (int i = 1; i < n; i++) {
c[i] = b[i] / a[i];
}
for (int i = 1; i < n; i++) {
for (int j = 1; j < n - i; j++) {
if (c[j] < c[j + 1]) {
double temp = c[j];
c[j] = c[j + 1];
c[j + 1] = temp;
double temp2 = a[j];
a[j] = a[j + 1];
a[j + 1] = temp2;
double temp3 = b[j];
b[j] = b[j + 1];
b[j + 1] = temp3;
}
}
}
}
static double Knapsack(double[] a, double[] b, double[] x, double C, int n) {
sort(a, b, n);
int i;
double total = 0;
for (i = 0; i < n; i++) {
if (a[i] <= C) {//如果揹包剩餘的容量大於等於第i個物體的重量
x[i] = 1; //把第i個物體整個裝進揹包
C = C - a[i]; //揹包的剩餘容量減少了第i個物體的重量
} else {
break; //退出循環
}
}
if (i < n) {//判斷是否第n個物體整個裝進去揹包裏了,如果i<=n表示否定
x[i] = C / a[i];
}
for (i = 0; i < n; i++) {
total = total + x[i] * b[i];
}
return total;
}
}