题目描述:
有编号分别为a,b,c,d,e的五件物品,它们的重量分别是2,2,6,5,4,它们的价值分别是6,3,5,4,6,现在给你个承重为10的背包,如何让背包里装入的物品具有最大的价值总和?
| name | weight | value | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| a | 2 | 6 | 0 | 6 | 6 | 9 | 9 | 12 | 12 | 15 | 15 | 15 |
| b | 2 | 3 | 0 | 3 | 3 | 6 | 6 | 9 | 9 | 9 | 10 | 11 |
| c | 6 | 5 | 0 | 0 | 0 | 6 | 6 | 6 | 6 | 6 | 10 | 11 |
| d | 5 | 4 | 0 | 0 | 0 | 6 | 6 | 6 | 6 | 6 | 10 | 10 |
| e | 4 | 6 | 0 | 0 | 0 | 6 | 6 | 6 | 6 | 6 | 6 | 6 |
public class Main {
public static void main(String[] args) {
// TODO Auto-generated method stub
final int packageWheight=10;//包的重量
Package[] pg={
new Package(6,2,"a"),
new Package(3,2,"b"),
new Package(5,6,"c"),
new Package(4,5,"d"),
new Package(6,4,"e")
};
int[][] bestValues = new int[pg.length+1][packageWheight+1];
for(int i=0;i<=pg.length;i++){
for(int j=0;j<=packageWheight;j++){
if(i==0||j==0){
bestValues[i][j]=0;//临界情况
}
else{
if(j<pg[i-1].getWheight()){
bestValues[i][j] = bestValues[i-1][j];//当第n件物品重量大于包的重量时,最佳值取前n-1件的
}
else{
int iweight = pg[i-1].getWheight(); //当第n件物品重量小于包的重量时,分两种情况,分别是装第n件或不装,比较取最大
int ivalue = pg[i-1].getValue();
bestValues[i][j] =
Math.max(bestValues[i-1][j], ivalue + bestValues[i-1][j-iweight]);
}
}
}
}
System.out.print(""+bestValues[pg.length][packageWheight]);
}
}
public class Package {
int value;
int wheight;
String name;
Package(int value,int wheight,String name){
this.value=value;
this.wheight=wheight;
this.name=name;
}
public int getWheight(){
return wheight;
}
public int getValue(){
return value;
}
public String getName(){
return name;
}
}
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* 本文来自博客 “李博Garvin“
* 转载请标明出处:http://blog.csdn.net/buptgshengod
******************************************/
【算法数据结构Java实现】Java实现动态规划(背包问题)
原文:http://blog.csdn.net/buptgshengod/article/details/43529417