c - 背包-节省时间和内存

根据维基百科和我查过的其他资料，你需要矩阵m[n][W]；n-物品的数量和背包的总容量。这个矩阵变得非常大，有时太大，无法在C程序中处理我知道动态编程是基于节省内存时间的，但是，还有什么解决方案可以节省时间和内存吗？
Knapsack problem:的伪代码

// Input:
// Values (stored in array v)
// Weights (stored in array w)
// Number of distinct items (n)
// Knapsack capacity (W)
for j from 0 to W do
  m[0, j] := 0
end for
for i from 1 to n do
  for j from 0 to W do
    if w[i] <= j then
      m[i, j] := max(m[i-1, j], m[i-1, j-w[i]] + v[i])
    else
      m[i, j] := m[i-1, j]
    end if
  end for
end for

假设W=123456789，n=100在这种情况下，我们得到真正大的矩阵m[100][123456789]我在想如何实现这一点，但我脑子里想的最好的办法是只保存用一位（0/1）选择的项。这可能吗或者有其他方法解决这个问题吗？

int32 -> 32 * 123456789 * 100 bits
one_bit -> 1 * 123456789 * 100 bits

我希望这不是愚蠢的问题，谢谢你的努力。
编辑-工作C代码：

    long int i, j;
    long int *m[2];
    m[0] = (long int *) malloc(sizeof(long int)*(W+1));
    m[1] = (long int *) malloc(sizeof(long int)*(W+1));
    for(i = 0; i <= W; i++){
        m[0][i] = 0;
    }

    int read = 0;
    int write = 1;
    int tmp;

    long int percent = (W+1)*(n)/100;
    long int counter = 0;

    for(i = 1; i <= n; i++){
        for(j = 0; j <= W; j++){
            if(w[i-1] <= j){
                m[write][j] = max(m[read][j],(v[i-1]) + m[read][j-(w[i-1])]);
            }else{
                m[write][j] = m[read][j];
            }
            counter++;
            if(counter == percent){
                printf(".");    //printing dot (.) for each percent
                fflush(stdout);
                counter = 0;
            }
        }
        tmp = read;
        read = write;
        write = tmp;
    }

    printf("\n%ld\n", m[read][W]);

    free(m[0]);
    free(m[1]);

最佳答案

背包问题可以用O(W)空间来求解。
在迭代的每个步骤中，您只需要2行-数组的当前状态m[i]和m[i + 1]。

current = 1
int m[2][W]
set NONE for all elements of m # that means we are not able to process this state
m[0][0] = 0 # this is our start point, initially empty knapsack

FOR i in [1..n] do
    next = 3 - current; /// just use 1 or 2 based on the current index
    for j in [0...W] do
       m[next][j] = m[current][j]
    FOR j in [w[i]..W] do
       if m[current][j - w[i]] is not NONE then  # process only reachable positions
           m[next][j] = max(m[next][j], m[current][j - w[i]] + v[i]);
    current = next; /// swap current state and the produced one

也可以只使用一个数组。这是伪代码

FOR i in [1..n] do
    FOR j in [w[i]..W] do
       m[j] = max(m[j], m[j - w[i]] + v[i]);