算法java实现--回溯法--0-1背包问题

0-1背包问题的java实现（回溯法）

具体问题描述以及C/C++实现参见网址

http://blog.csdn.net/liufeng_king/article/details/8764319

/**
 * 0-1背包问题--回溯法
 * @author Lican
 *
 */
public class Knapsack {
	public static class Element implements Comparable{
		int id;//物品编号
		double d;
		public Element(int id,double d){
			this.id=id;
			this.d=d;
		}
		@Override
		public int compareTo(Object x) {
			double xd=((Element)x).d;//递减顺序排列
			if(dn){//到达叶子节点
			bestp=cp;
			for(int j=1;j<=n;j++){//保存最优值对应的包的编号
				bestx[j]=x[j];
			}
			return;
		}
		if(cw+w[i]<=c){//左子树
			x[i]=1;
			cw+=w[i];
			cp+=p[i];
			backtrack(i+1);
			cw-=w[i];//恢复现场
			cp-=p[i];
		}
		if(bound(i+1)>bestp){
			x[i]=0;
			backtrack(i+1);
		}
	}
	public double bound(int i){//上界函数
		double cleft=c-cw;
		double bound=cp;
		while(i<=n&&w[i]<=cleft){
			cleft-=w[i];
			bound+=p[i];
			i++;
		}
		if(i<=n){
			bound+=p[i]*cleft/w[i];
		}
		return bound;
	}
	public static void main(String[] args) {
		double[] weight={0,71,34,82,23,1,88,12,57,10,68,5,33,37,69,98,24,26,83,16,26,18,43,52,71,22,65,68,8,40,40,24,72,16,34,10,19,28,13,34,98,29,31,79,33,60,74,44,56,54,17};
		double[] price={0,26,59,30,19,66,85,94,8,3,44,5,1,41,82,76,1,12,81,73,32,74,54,62,41,19,10,65,53,56,53,70,66,58,22,72,33,96,88,68,45,44,61,78,78,6,66,11,59,83,48};
		double cc=300;
		//double[] weight={0,7,3,4,5};
		//double[] price={0,42,12,40,25};
		//double cc=10;
		Knapsack k=new Knapsack();
		double best=k.knapsack(price,weight,cc);
		System.out.println("最优值："+best);
		System.out.println("选中的物品编号分别是：");
		for(int i=1;i