C + + realizes steel strip cutting problem

Looking at the dynamic programming (DP) in the introduction to the algorithm today, there is an interesting question: the steel strip cutting to get the maximum benefit.

Several solutions are discussed, including recursive solution method, Memo DP solution, bottom-up solution and the reconstruction of the solution.
The book gives the pseudo code of different solutions, just need to practice C + +, there is C + + to achieve DP solution to the steel bar cutting problem.

"Recursive solver"

Steel bar cutting problem//top-down recursive implementation # include <iostream> #include <time.h>using namespace std;int cut_rod (int len, int price_a    Rr[]); int main (int argc, char *argv[]) {clock_t start, finish;    Double duration;    start = Clock ();    int rod_len = 10;    int p_arr[] = {0, 1, 5, 8, 9, 10, 17, 17, 20, 24, 30};    int profit = Cut_rod (Rod_len, P_arr);    cout << "Profit =" << profit << Endl;    finish = Clock ();  Duration = (double) (Finish-start)/clocks_per_sec;    Units s cout << "time Cost =" << duration * + << "MS" << Endl; return 0;} /* Recursive call to Cut_rod function * Input: Rod length len, Unit bar length Price table price_arr[] * Output: Len's best yield for steel bars * * thought: * 1. If len==0, return 0 * 2. Otherwise, the steel bar is cut to the left length of I and the right length of len-i two segments, * length of I is not in the cut, the profit is Price_arr[i], the right part continues to decompose. * Note: * 1. There is an access risk to the program, and when the input length is greater than p_arr[] length, the elements outside the array are accessed; * Therefore the program is valid only for the data of Len < sizeof (P_arr)/sizeof (int); */int cut_rod (int len, int p    Rice_arr[]) {if (len = = 0) return 0;    int best_profit =-100; for (iNT i = 1; I <= Len;    i++) {best_profit = max (Best_profit, Price_arr[i] + cut_rod (len-i, Price_arr)); } return best_profit;}

"Bottom-up DP reconstruction Solution"

The bottom-up reconstruction solution includes the bottom-up solution, so that only the reconstruction method is transmitted here.

The article assumes that the price is still $30 when the bar length is greater than 10 inches, so there is no limit to the length of the input bar < of course it is unscientific >;

/* DP solves the problem of strip cutting * Not only returns the maximum yield of the steel bar with len length r[], and returns the cutting condition s[] (the solution of the refactoring) * The solution of the reconstruction is also determined by the nature of the sub-problem. */#include <iostream> #include < string.h> #define Rod_len 17#define len Rod_len + 1using namespace std;void extend_bottom_up_cut_rod (int len, int price_  arr[], int (&r) [Len], int (&s) [Len]), void print_cut_rod_solution (int LEN, int price_arr[], int (&r) [Len], int (&s) [LEN]);    int main (int argc, char *argv[]) {int price[] = {0, 1, 5, 8, 9, 10, 17, 17, 20, 24, 30};    int R[len], S[len];    memset (r, 0, sizeof (r));    memset (s, 0, sizeof (s));    Print_cut_rod_solution (Rod_len, Price, R, s); return 0;} void Extend_bottom_up_cut_rod (int len, int price_arr[], int (&r) [Len], int (&s) [Len]) {if (len = = 0) ret    Urn        for (int j = 1; j <= Len; j + +) {int p =-1000;            for (int i = 1; I <= J; i++)//J-Length steel strip, cut i {int left_price;            if (i >) left_price = price_arr[10]; else Left_price = Price_arr[i];                if (P < Left_price + r[j-i]) {p = left_price + r[j-i];            S[J] = i;    }} R[j] = p; }}void print_cut_rod_solution (int len, int price_arr[], int (&r) [Len], int (&s) [Len]) {Extend_bottom_up_cut_rod    (Len, Price_arr, R, s);    cout << len << "inch rod Price is" << R[len] << Endl;    cout << "The Cut Order is";        while (Len > 0) {cout << S[len] << "";    Len-= S[len]; }}

　　

C + + realizes steel strip cutting problem