Introduction to Algorithms 15.1 steel strip cutting

The implementation of this DP problem is implemented in code, which is divided into recursive algorithm (top-down) and non-recursive algorithm (from bottom-up), and the implementation of the extended bottom-up algorithm.

Recursive algorithm:

1#include <iostream>2 3 using namespacestd;4 5 intSize =Ten;6 7InlineintMaxintAintb)8 {9     returna > B?a:b;Ten } One  A intCutrod (intP[],intN) - { -     if(n = =0) the         return 0; -     Else -     { -         intQ =int_min; +          for(inti =1; I < n; i++) -         { +Q = max (q, P[i] + Cutrod (p, N-i-1)); A         } at         returnQ; -     } - } -  -  -  in intMain () - { to     intA[] = {1,5,8,9,Ten, -, -, -, -, - }; +cout << Cutrod (A,6) <<Endl; -}

Non-recursive algorithm:

#include <iostream>using namespacestd;intSize =Ten; inlineintMaxintAintb) {    returna > B?a:b;}intNonrecursioncutrod (intP[],intN) {    int*r =New int[n]; r[0] = p[0];  for(intj =1; J < N; J + +)    {        intQ =P[j];  for(inti =0; I < J; i++)        {            if(q< (P[i] + r[j-i-1])) {Q= P[i] + r[j-i-1]; }} R[j]=Q; }    returnr[n-1];}intMain () {intA[] = {1,5,8,9,Ten, -, -, -, -, - }; cout<< Nonrecursioncutrod (A,6) <<Endl;}

Extended non-recursive algorithm:

Here I set up a result struct to be used as the return results, in fact this intention: if we ask for a solution of length J, then we will put the length of the first segment of the cut in the cell labeled J in the array, if I, then the remainder can be found by looking for the j-i scheme, This way, a solution of length J can be obtained by recursion.

1#include <iostream>2 3 using namespacestd;4 5typedefstructResult6 {7     intMax;8     int*Price ;9 }result;Ten  OneResult Extendnonrecursioncutrod (intA[],intN) A { - result result; -Result. Price =New int[n+1]; the     int*r =New int[n+1]; -r[1] = a[1]; -Result. price[1] = a[1]; -      for(intj =1; J <= N; J + +) +     { -         intQ =A[j]; +         intt =J; A          for(inti =1; I < J; i++) at         { -             if(Q < (A[i] + r[j-i])) -             { -Q = a[i] + r[j-i]; -t =i; -             } in         } -R[J] =Q; toResult. PRICE[J] =T; +     } -Result. Max =R[n]; the     returnresult; * } $ Panax Notoginseng intMain () - { the     intA[] = {0,1,5,8,9,Ten, -, -, -, -, - }; +Result result = Extendnonrecursioncutrod (A,Ten); Acout << result. Max <<Endl; the      for(inti =1; I <Ten; i++) +cout << result. Price[i] <<"    "; -}

Introduction to Algorithms 15.1 steel strip cutting