O-1 knapsack - dynamic programming

/*
    Dynamic programming approach to 0-1 knapsack problem - O(mn)
    Two cases arises for each  nd every item - include or discard - 0-1
     n This should be the f$ckin rule of life too :P

    Max value is obtained from n items is max of the  following -

    1 - max value obtained from n-1 items with W as weight
    (excluding nth item).

    2- value of nth item plus max value oobtained by n-1 items and weight as (W - weight(nth) ) item 
    (including nth item)

    If weight of nth > W then only case left is 1st , 0-1 hence

*/

#include<stdio.h>

int max(int a, int b) { return (a > b) ? a : b; };

//returns max value to be sacked in a knapsack of capacity W
int knapSack(int W,int wt[], int val[], int n)
{
    int i, w;
    int K[n + 1][W + 1];

    //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;
            //include
            else if (wt[i - 1] <= w)
                K[i][w] = max(val[i - 1] + K[i - 1][w - wt[i - 1] ], K[i - 1][w]);
            else
                //discard
                K[i][w] = K[i - 1][w];
        }
    }

    return K[n][W];
}


int main()
{
    int val[] = { 60, 100, 120 };
    int wt[] = { 10, 20, 30 };
    int  W = 50;
    int n = sizeof(val) / sizeof(val[0]);
    printf("%d", knapSack(W, wt, val, n));
    getchar();
    return 0;
}