比如:
1.有两艘船,载重量分别是w1,和w2,现在有一批货,如何装载,是的两艘船能够装载尽量多的货物。
这个题目就可以用01背包的方法解决。即先计算一艘船的背包问题,然后用剩下的货物计算另一艘船的背包问题。
2.给定一个正整数的集合和一个值(c),是否存在集合的一个子集,是的子集里元素的和是否等于c。
与01背包的解法类似。
3.设某一机器有n个部件组成,每一种部件都可以从m个不同的的供应商出购得,各种供应商提供的相同部件的价格和重量都不同,是找出总价格不超过价格上限的最小机器重量.
这个问题同样可以用01背包解决,只是在回溯函数里,需要加入一个循环,计算每个供应商购得零件后的价格和重量.
/**
* 01背包问题
* 使用回溯算法实现
* @author ztl
*
*/
public class Knapsack {
private int[] weight;
private int maxWeight;
private int currentWeight;
private int bestWeight;
private int height;
private int[] state;
private int[] currentState;
public Knapsack(int mw,int[] weightSet){
this.maxWeight = mw;
this.weight = weightSet;
this.currentWeight = 0;
this.bestWeight = 0;
this.height = weightSet.length - 1;//要减一,是因为第一个节点的height是0
state = new int[weightSet.length];
currentState = new int[weightSet.length];
backtrace(0);
printResult();
}
public void backtrace(int i){
if (i > height){//到达了叶节点
if (currentWeight > bestWeight){
bestWeight = currentWeight;
for (int j=0;j<state.length;j++){
state[j] = currentState[j];
}
}
return;
}
if (currentWeight + weight[i] <= maxWeight){//检查右子树
currentWeight += weight[i];
currentState[i] = 1;
backtrace(i+1);
currentWeight -= weight[i];
currentState[i] = 0;
}
backtrace(i+1);//检查左子树
}
public void printResult(){
System.out.println("result is "+bestWeight);
for (int i=0;i<state.length;i++){
System.out.print(state[i] + " ");
}
System.out.println();
}
public static void main(String args[]){
int[] weight = {2,5,4,7,16,11,13};
new Knapsack(10,weight);
}
}