int memoized_cut_rod_aux(int *p, int n, int *r) {
int q;
if (r[n] >= 0) {
return r[n];
}
if (n == 0) {
q = 0;
} else {
q = INT_MIN;
for (int i = 0; i < n; i++) {
int j = i > 9 ? 9 : i;
q = MAX(q, p[j] + memoized_cut_rod_aux(p, n - i - 1, r));
}
}
r[n] = q;
printf("r = ");
for (int i = 0; i < n; i++) {
printf("%d ", r[i]);
}
printf("\n");
return q;
}
/*
* return the max proceed of n-length rod,
* n is not equal the whole rod_length
*/
int memoized_cut_rod_aux( int *p, int n, int *s, int *r )
{
/* r[n] is the memo */
if( r[n] >= 0 )
return r[n];
int q = INT_MIN;
if( n == 0 ){
q = 0;
} else {
/*
* i counts from 1 but not 0,
* because a function can't call itself with the same parameters.
*/
for( int i = 1; i <= n; i++ ){
int current_proceed = p[i]
+ memoized_cut_rod_aux( p, n-i, s, r );
if( q < current_proceed ){
q = current_proceed;
s[n] = i;
}
}
}
return q;
}
/*
* memoized method, form top downto bottom,
* calculate the necessary the subquestion only
* and store it for further uses.
*/
int memoized_cut_rod( int *p, int n, int *s )
{
int *r = (int *)malloc( sizeof(int) * ( n + 1 ));
for( int i = 0; i <= n; i++ )
r[i] = INT_MIN;
return memoized_cut_rod_aux( p, n, s, r );
}
int memoized_cut_rod(int *p, int n) {
int r[n+1];
for (int i = 0; i < n + 1; i++) {
r[i] = INT_MIN;
}
return memoized_cut_rod_aux(p, n, r);
}
