问题
I know how to solve knapsack 0-1 problem with dynamic programming approach, but I am having troubles figuring out which items to take without compromising the complexity of O(N * C) (N items, C capacity).
Any ideas (I would prefer a bottom-up approach)?
回答1:
Suppose, right now you're storing results in array bool[] a
, where a[i]
is true when sum i
can be achieved.
You'll need another array int[] b
, where b[i]
is a last element you've placed into knapsack to achieve sum i
.
So, where you had
a[i] = true;
you'll need
a[i] = true;
b[i] = current_item;
Then, finding which items can be taken to achieve sum i
is a simple loop.
PS I use two arrays for simplicity, but obviously array a
can be removed.
回答2:
Here is a modification to reconstruct path in O(n) times
int knapsack(int weight[], int profit[], int no_of_items, int capacity) {
for (int var = 0; var <= capacity; ++var) {
dp[0][var] = 0;
}
for (int var = 0; var <= no_of_items; ++var) {
path[var] = false;
}
int using_item_i, without_using_item_i;
for (int i = 1; i <= no_of_items; ++i) {
for (int j = 1; j <= capacity; ++j) {
without_using_item_i = dp[i - 1][j];
using_item_i = 0;
if ((weight[i]) <= j) {
using_item_i = dp[i - 1][j - weight[i]] + profit[i];
}
if (using_item_i >= without_using_item_i) {
taken[i][j] = true;
dp[i][j] = using_item_i;
} else {
taken[i][j] = false;
dp[i][j] = without_using_item_i;
}
}
}
//Reconstructing back the path
int j = capacity;
for (int i = no_of_items; i >= 0; --i) {
if (taken[i][j]) {
path[i] = true;
cnt++;
}
j = j - weight[i];
}
return dp[no_of_items][capacity];
}
回答3:
boolean[] solution = new boolean[nItems];
for (int i = nItems, c = maxCapacity; i > 0 && c > 0; i--) {
int iThItemAddedValue = value[i - 1][c - weights[i - 1]] + values[i - 1];
int iThItemInheritedValue = value[i - 1][c];
if (iThItemAddedValue > iThItemInheritedValue) {
solution[i - 1] = true;
c = c - weights[i - 1];
} else {
solution[i - 1] = false;
}
}
回答4:
Check the sol in the attached image
回答5:
public class Knapsackproblem {
private static int[][] cache;
public static void main(String[] args) {
int val[] = new int[]{60, 100, 120};
int wt[] = new int[]{10, 20, 30};
int W = 50;
int n = val.length;
System.out.println(knapSack(W, wt, val, n));
printValues(wt,val);
}
/**
* This method will find the result with
* more value with weight less than or equal
* to given weight
* @param w given weight
* @param wt arrays of weights
* @param val array of values
* @param n length of the array
* @return max value we can obtain
*/
private static int knapSack(int w, int[] wt, int[] val, int n) {
cache = new int[n+1][w+1];
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= w; j++) {
if(j < wt[i-1]){
cache[i][j] = cache[i-1][j];
}else {
cache[i][j] = Math.max(cache[i-1][j],(cache[i-1][j-wt[i-1]])+val[i-1]);
}
}
}
for (int[] aCache : cache) {
System.out.println(Arrays.toString(aCache));
}
return cache[n][w];
}
private static void printValues(int[] wt, int[] val) {
int m = cache.length-1;
int n = cache[0].length-1;
util(wt,val,m,n);
}
private static void util(int[] wt, int[] val, int m, int n) {
if(m <=0 || n<=0) return;
if((cache[m][n] != cache[m-1][n]) && (cache[m][n] != cache[m][n-1])){
System.out.println(val[m-1]+"-->"+wt[m-1]);
util(wt, val, m-1, (n - wt[m - 1] + 1));
}else
if(cache[m][n] == cache[m-1][n]){
util(wt,val,m-1,n);
}
else if(cache[m][n] == cache[m][n-1])
util(wt,val,m,n-1);
else
util(wt,val,m,(n-val[m-1]+1));
}
}
回答6:
https://www.dropbox.com/s/ish7t5vgy91fovt/Screenshot%202017-01-01%2015.16.31.png?dl=0
Print the tmpList in the caller and you will get the answer
来源:https://stackoverflow.com/questions/4935533/knapsack-0-1-path-reconstruction-which-items-to-take