Change-making problem

http://en.wikipedia.org/wiki/Change-making_problem

 

Given unlimited amounts of coins of denominations d1 > … > dm , must contain 1.
give change for amount n with the least number of coins.
Does the greedy algorithm always give an optimal solution for the general change-making problem We give the following example,
Example: d1 = 7c, d2 =5c, d3 = 1c, and n = 10c, not always produces an optimal solution. in fact, for some instances, the problem may not have a solution at all! consider instance d1 = 7c, d2 =5c, d3 = 3c, and n = 11c.
But, this problem can be solved by dynamic programming, please try to design an algorithm for the general Change-making_problem after class.

该算法复杂度为O(NK)

 

#include <stdio.h>#include <tchar.h>#include <windows.h>#include <iostream>#include <fstream>#include <string> #include <sstream>#include <vector>using namespace std;/*!@brief Calculate the minimum number of stamps needed for the desired value.@details Use Dynamic Programing to solve this problem.@param[in] array Input array of sorted stamp values.@param[in] array_siz Number of elements in array.@param[in] request Desired value to of stamps.@return The minimum stamps needed for the desired value.@note Assume that there will always be a one cent stamp in the machine.*/int min_number_of_stamps(const int* array, size_t array_size, int request){if(array == NULL || array_size == 0 || request == 0) return 0;int* minStamp = (int*)malloc((request + 1) * 4); //minStamp[i] = the minimum stamps needed for i. Allocate space for all the numbers smaller then request.memset(minStamp,0,(request + 1) * 4);//initialize the space to 0, minStamp[0] = 0;int minimum;//the minimum stamps needed for request.//Assume MinNumStamp(i) returns the minimum stamps needed for i.////We can pick up array[j] (array[j] < i) as a temporary selected stamps, assume it's in the minimum needed stamps array.//Then the remain needed value is "i-array[j]",and the number of the remain needed stamps is MinNumStamp(i-array[j]).////The formula is:// MinNumStamp(i) = min { MinNumStamp(i-array[j]) + 1, array[j] <= i}////MinNumStamp(i-array[j]) is the sub-problem of MinNumStamp(i).//In the previous steps, after having calculated MinNumStamp(i-array[j]), we can store the result in minStamp[i-array[j]].//When we caculat the MinNumStamp(i), we can directly read minStamp[i-array[j]], so that don't need to re-calculate the MinNumStamp(i-array[j]).//It It's a typical Dynamic Programing problem.////With initial condition minStamp[0] = 0, we can caculate from i = 1 to i = request successively.//i-array[j] < i (for every array[j] <= i), so that when caculating MinNumStamp(i), all the MinNumStamp(i-array[j]) have already been stored in minStamp[i-array[j]].//So we can turn the formula to:// minStamp[i] = min { minStamp[i-array[j]] + 1, array[j] <= i}for(int i = 1; i <= request; i++)//caculate minimum stamps needed for i successively, and store it in minStamp[i].{int min = 2147483647;//2147483647 is the MAX_INT, to store MinNumStamp(i).int cur_min;//store MinNumStamp(i-array[j]) + 1.for(size_t j = array_size - 1; j != -1; j--)//choose array[j] as temporary selected stamps.{if(array[j] > i) break;//if array[j] > i, it must not be the stamp needed for i.cur_min = minStamp[i - array[j]] + 1;if(cur_min < min) min = cur_min; //minStamp[i] = min { minStamp[i-array[j]] + 1, array[j] <= i}}minStamp[i] = min; }minimum = minStamp[request];free(minStamp);return minimum;}int _tmain(int argc, _TCHAR* argv[]){cout<<"Please specify the input file which include input array of sorted stamp values."<<endl;cout<<"If the file dosen't exit, we will use the default testing data."<<endl;vector<int> vStampValues;string sFileName;getline(cin, sFileName);ifstream infile(sFileName.c_str());if(!infile){cout<<"The file dosen't exit, we will use the default testing data."<<endl;string sInput = "1001 100 90 30 24 15 12 10 5 3 2 1";istringstream iss(sInput);int i;while(iss >> i){vStampValues.push_back(i);}}else{int i;while(infile>>i){vStampValues.push_back(i);}}const int* const array = reinterpret_cast<int*>(&vStampValues.at(0));cout<<"Please input the value request to calculate."<<endl;int request;while(cin>>request){DWORD t1 = GetTickCount();cout<<min_number_of_stamps(array, vStampValues.size(), request)<<endl;cout<<"Time:"<<GetTickCount() - t1<<endl;}if(infile){infile.close();}return 0;}