回溯法解0-1背包问题

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回溯法：

Knapsack.java

public class Knapsack {private double c;         //背包重量private int n;            //物品数量private double[] w;       //物品重量private double[] v;       //物品价值private int[] num;        //物品序号private double cw;        //当前背包重量private double cv;        //当前背包价值private double maxv;      //当前最优值private int[] bestx;      //当前最优解private int[] x;          //部分解public Knapsack(double[] weight, double[] value, double c){this.c =c;this.n = weight.length;w = new double[n];v = new double[n];num = new int[n];for(int i = 0; i < n; i++) {num[i] = i + 1;w[i] = weight[i];v[i] = value[i];}//将各物品按单位重量价值从大到小排序（排序）double tmp;int tmpp;for(int i = 0; i < n; i++) {for(int j = 1; j < n; j++) {if((v[j] / w[j]) > (v[j-1] / w[j-1])) {tmp = v[j-1];v[j-1] = v[j];v[j] = tmp;tmp = w[j-1];w[j-1] = w[j];w[j] = tmp;tmpp = num[j-1];num[j-1] = num[j];num[j] = tmpp;}}}}void cacu() {cw = 0;cv = 0;maxv = 0;bestx = new int[n];x = new int[n];backtrack(0);System.out.print("装入背包的总价值：");System.out.println(maxv);System.out.print("装入背包的物品序号：");for(int i=0; i < n; i++) {if(bestx[i] == 1) {System.out.print(num[i] + " ");}}}void backtrack(int t) {if(t >= n) {output(x);return;}for(int i=1; i>=0; i--) {x[t] = i;if(i == 1) {cw += w[t];cv += v[t];if(constraint(t)&&bound(t)) {backtrack(t+1);}cw -= w[t];cv -= v[t];}if(i == 0) {if(constraint(t)&&bound(t)) {backtrack(t+1);}}}}void output(int[] x) {for(int i = 0; i < n; i++) {bestx[i] = x[i];}maxv = cv;}boolean constraint(int t) {if(c < cw)  return false;return true;}boolean bound(int t) {double cwr = cw;double cvr =cv;while(t < n-1 && (w[t+1] <= (c-cwr))) {cwr += w[t+1];cvr += v[t+1];t++;}if(cvr <= maxv)  return false;return true;}}

Test.java

import java.util.Scanner;public class Test {public static void main(String[] args) {double c = 50;int n = 0;Scanner sc = new Scanner(System.in);System.out.print("输入背包容量：");c = sc.nextDouble();System.out.print("输入物品数量：");n = sc.nextInt();double[] w = new double[n];double[] v = new double[n];for(int i=1; i<=n; i++){System.out.print("输入物品" + i + "质量和价值：");w[i-1] = sc.nextDouble();v[i-1] = sc.nextDouble();}Knapsack x = new Knapsack(w, v, c);long start = System.nanoTime();x.cacu();long end = System.nanoTime();System.out.println();System.out.println("花费时间: " + String.valueOf(end - start) + "纳秒");}}

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