如何使用动态规划在背包问题的最优解中找到项目？

Question

我想找到使用动态编程方法在背包问题的最佳解决方案中最终选择了哪些项目。
到目前为止，这是我的解释...

#include<stdio.h>

int getMax(int x, int y) {
    if(x > y) {
        return x;
    } else {
        return y;
    }
}

int main(void) {

    //the first element is set to -1 as
    //we are storing item from index 1
    //in val[] and wt[] array
    int val[] = {-1, 100, 20, 60, 40};  
    int wt[] = {-1, 3, 2, 4, 1};   
    int A[] = {0,0,0,0,0};    
    int n = 4; //num
    int W = 5;//cap  
    int i, j;


    // value table having n+1 rows and W+1 columns
    int V[n+1][W+1];

    // fill the row i=0 with value 0
    for(j = 0; j <= W; j++) {
        V[0][j] = 0;
    }

    // fill the column w=0 with value 0
    for(i = 0; i <= n; i++) {
        V[i][0] = 0;
    }

    //fill the value table
    for(i = 1; i <= n; i++) {
        for(j = 1; j <= W; j++) {
            if(wt[i] <= j) {
                V[i][j] = getMax(V[i-1][j], val[i] + V[i-1][j - wt[i]]);
            } else {
                V[i][j] = V[i-1][j];
            }
        }
    }

    //max value that can be put inside the knapsack
    printf("Max Value: %d\n", V[n][W]);


    //==================================find items
    int n1,c;
    n1=n;
    c=W;
    int A2[n1][c];

    while(c>0){

        if(A2[n1][c]==A2[n1-1][c]){
            A[n1]=0;
        } else {
            A[n1]=1;
        }

        n1=n1-1;
        c=c-wt[n1];

    }

    printf("Final array of items: ");
    for(i = 0; i < n; i++){
        printf("%d",A[i]);
    }
} // end of main

这是输出：

Max Value: 140
Final array of items: 0001

这串一和零应该是最终选择的项目，但是从解决方案来看这似乎是错误的！

我遵循此算法：

While the remaining capacity is greater than 0 do
If Table[n, c] = Table[n-1, c] then
Item n has not been included in the optimal solution
Else
Item n has been included in the optimal solution
Process Item n
Move one row up to n-1
Move to column c – weight(n)

那么，此算法是否错误/不适合该方法，或者我遗漏了什么？