二维数组中未处理的异常错误

Question

这个动态规划算法返回未处理的异常错误，可能是由于我用于各种(非常大)输入数量的二维数组。我似乎不明白这是怎么回事。完整的程序如下:

// A Dynamic Programming based solution for 0-1 Knapsack problem
#include<stdio.h>
#include<stdlib.h>
#define MAX 10000

int size;
int Weight;
int p[MAX];
int w[MAX];
// A utility function that returns maximum of two integers
int maximum(int a, int b) { return (a > b) ? a : b; }
// Returns the maximum value that can be put in a knapsack of capacity W
int knapSack(int W, int wt[], int val[], int n)
{
int i, w;
int retVal;
int **K;
K = (int**)calloc(n+1, sizeof(int*));
for (i = 0; i < n + 1; ++i)
{
    K[i] = (int*)calloc(W + 1, sizeof(int));
}
// Build table K[][] in bottom up manner
for (i = 0; i <= n; i++)
{
    for (w = 0; w <= W; w++)
        {
    if (i == 0 || w == 0)
      K[i][w] = 0;
            else if (wt[i - 1] <= w)
      K[i][w] = maximum(val[i - 1] + K[i - 1][w - wt[i - 1]], K[i - 1][w]);
            else
          K[i][w] = K[i - 1][w];
    }
}
retVal = K[n][W];
for (i = 0; i < size + 1; i++)
    free(K[i]);
free(K);
return retVal;
}
int random_in_range(unsigned int min, unsigned int max)
{
int base_random = rand();
if (RAND_MAX == base_random) return random_in_range(min, max);
int range = max - min,
    remainder = RAND_MAX % range,
    bucket = RAND_MAX / range;
if (base_random < RAND_MAX - remainder) {
    return min + base_random / bucket;
}
else {
    return random_in_range(min, max);
}
}
int main()
{
srand(time(NULL));
int val = 0;
int i, j;
//each input set is contained in an array
int batch[] = { 10, 20, 30, 40, 50, 5000, 10000 };
int sizeOfBatch = sizeof(batch) / sizeof(batch[0]);
//algorithms are called per size of the input array
for (i = 0; i < sizeOfBatch; i++){
    printf("n");
    //dynamic array allocation (variable length to avoid stack overflow
    //calloc is used to avoid garbage values
    int *p = (int*)calloc(batch[i], sizeof(int));
    int *w = (int*)calloc(batch[i], sizeof(int));
    for (j = 0; j < batch[i]; j++){
        p[j] = random_in_range(1, 500);
        w[j] = random_in_range(1, 100);
    }
    size = batch[i];
    Weight = batch[i] * 25;
    printf("|   %d ", batch[i]);
    printf("   %d", knapSack(Weight, w, p, size));
    free(p);
    free(w);
}
_getch();
return 0;
}

Answer 1

改变:

for (i = 0; i < size + 1; i++)
    free(K[i]);
free(K);
return K[size][Weight];

:

int retVal;
...
retVal = K[size][Weight];
for (i = 0; i < size + 1; i++)
    free(K[i]);
free(K);
return retVal;