【发布时间】：2011-09-19 17:38:37
【问题描述】：
我基于pseudo-Java code from here 编写了分支定界背包算法的实现。不幸的是，在问题的大型实例like this 上，内存阻塞。为什么是这样？我怎样才能使这个实现更有效地使用内存？
链接上文件的输入格式如下：
numberOfItems maxWeight
profitOfItem1 weightOfItem1
.
.
.
profitOfItemN weightOfItemN
// http://books.google.com/books?id=DAorddWEgl0C&pg=PA233&source=gbs_toc_r&cad=4#v=onepage&q&f=true
import java.util.Comparator;
import java.util.LinkedList;
import java.util.PriorityQueue;
class ItemComparator implements Comparator {
public int compare (Object item1, Object item2){
Item i1 = (Item)item1;
Item i2 = (Item)item2;
if ((i1.valueWeightQuotient)<(i2.valueWeightQuotient))
return 1;
if ((i2.valueWeightQuotient)<(i1.valueWeightQuotient))
return -1;
else { // costWeightQuotients are equal
if ((i1.weight)<(i2.weight)){
return 1;
}
if ((i2.weight)<(i1.weight)){
return -1;
}
}
return 0;
}
}
class Node
{
int level;
int profit;
int weight;
double bound;
}
class NodeComparator implements Comparator {
public int compare(Object o1, Object o2){
Node n1 = (Node)o1;
Node n2 = (Node)o2;
if ((n1.bound)<(n2.bound))
return 1;
if ((n2.bound)<(n1.bound))
return -1;
return 0;
}
}
class Solution {
long weight;
long value;
}
public class BranchAndBound {
static Solution branchAndBound2(LinkedList<Item> items, double W) {
double timeStart = System.currentTimeMillis();
int n = items.size();
int [] p = new int [n];
int [] w = new int [n];
for (int i=0; i<n;i++){
p [i]= (int)items.get(i).value;
w [i]= (int)items.get(i).weight;
}
Node u;
Node v = new Node(); // tree root
int maxProfit=0;
int usedWeight=0;
NodeComparator nc = new NodeComparator();
PriorityQueue<Node> PQ = new PriorityQueue<Node>(n,nc);
v.level=-1;
v.profit=0;
v.weight=0; // v initialized to -1, dummy root
v.bound = bound(v,W, n, w, p);
PQ.add(v);
while(!PQ.isEmpty()){
v=PQ.poll();
u = new Node();
if(v.bound>maxProfit){ // check if node is still promising
u.level = v.level+1; // set u to the child that includes the next item
u.weight = v.weight + w[u.level];
u.profit = v.profit + p[u.level];
if (u.weight <=W && u.profit > maxProfit){
maxProfit = u.profit;
usedWeight = u.weight;
}
u.bound = bound(u, W, n, w, p);
if(u.bound > maxProfit){
PQ.add(u);
}
u = new Node();
u.level = v.level+1;
u.weight = v.weight; // set u to the child that does not include the next item
u.profit = v.profit;
u.bound = bound(u, W, n, w, p);
if(u.bound>maxProfit)
PQ.add(u);
}
}
Solution solution = new Solution();
solution.value = maxProfit;
solution.weight = usedWeight;
double timeStop = System.currentTimeMillis();
double elapsedTime = timeStop - timeStart;
System.out.println("* Time spent in branch and bound (milliseconds):" + elapsedTime);
return solution;
}
static double bound(Node u, double W, int n, int [] w, int [] p){
int j=0; int k=0;
int totWeight=0;
double result=0;
if(u.weight>=W)
return 0;
else {
result = u.profit;
totWeight = u.weight; // por esto no hace
if(u.level < w.length)
{
j= u.level +1;
}
int weightSum;
while ((j < n) && ((weightSum=totWeight + w[j])<=W)){
totWeight = weightSum; // grab as many items as possible
result = result + p[j];
j++;
}
k=j; // use k for consistency with formula in text
if (k<n){
result = result + ((W - totWeight) * p[k] / w[k]);// grab fraction of excluded kth item
}
return result;
}
}
}
【问题讨论】：
-
试着问一个更简洁的问题。真的很难理解这 100 多个 LOC 中的坏处。
标签： java optimization knapsack-problem