Introduction to C + + implementation algorithm 15 chapter of the Steel bar segmentation in dynamic programming

#include <iostream> #include <algorithm> #include <utility> #include <vector>using namespace std;//uses a common recursive algorithm to solve the largest yield of the steel bar segmentation int Cut_rod (int *p,const int &n) {if (n==0) return 0;int q=-1;for (int i=1;i<=n;++i) {Q =max (Q,p[i]+cut_rod (P,n-i));} return q;} A dynamic programming function method with a top-down function to solve the problem int memoized_cut_rod_aux (int *p,int n,int *r) {if (r[n]>=0) {return r[n];} int q=-1;if (n==0) q=0;else{for (int i=1;i<=n;++i) Q=max (Q,p[i]+memoized_cut_rod_aux (P,n-i,r));} R[n]=q;return Q;} Top-down main function int Memoized_cut_rod (int *p,int n) {int r[11]={};for (int i=0;i<11;++i) R[i]=-1;int q=memoized_cut_rod_ Aux (p,n,r); return q;} To the top-up method to solve the steel bar problem int Bottom_up_cut_rod (int *p,int n) {int r[11]={0};for (int j=1;j<=n;++j) {int q=-1;for (int i=1;i< =j;++i) {Q=max (q,p[i]+r[p,j-i]);} R[j]=q;}  return r[n];} pair< vector<int>,vector<int> > Extened_bottom_up__cut_rot (int *p,int n) {pair< vector<int  >,vector<int> > result;  int r[11]={0};  int s[11]={0};  Result.first.push_back (0);Result.second.push_back (0);  for (int j=1;j<=n;++j) {int q=-1;  for (int i=1;i<=j;++i) {//q=-1;  Q=max (Q,p[i]+r[j-i]) if (Q<p[i]+r[j-i]) {q=p[i]+r[j-i];  S[j]=i;  }} r[j]=q;  Result.first.push_back (R[j]);  Result.second.push_back (S[j]);  } return result;    }//print result value, void print_cut_rod_solution (int *p,int n) {cout<< "Maximum return value is" <<endl;  cout<< "Steel bars are split in the way" <<endl;  pair< vector<int>,vector<int> > result;  Result=extened_bottom_up__cut_rot (P,n);  cout<<result.first[n]<<endl;  cout<< "Steel bars are split in the way" <<endl;  while (n>0) {//cout<< "steel bars are split in the form of" <<endl;  cout<<result.second[n];  N=n-result.second[n];  cout<<endl;  } cout<<endl;  cout<< "End of the split of Steel bar" <<endl; }int Main () {int Price[11]={0,1,5,8,9,10,17,17,20,24,30};//cout<<cut_rod (price,5) <<endl;cout<< Bottom_up_cut_rod (price,8) <<endl;print_cut_rod_solution (price,8); System ("pause"); return 0;}

Introduction to C + + implementation algorithm 15 chapter of the Steel bar segmentation in dynamic programming