最终项目，动态规划。需要第二组眼睛
我觉得非常愚蠢的是今天在这里提出这个问题之后，在了解算法的每个人都是错误的。但我不再直视这件事了。无论如何，这是一个背包问题，通过memoization和动态编程解决。问题是我的答案的打印输出与要求不符。 我想要的只是第二次看它，如果有人可以指出我在哪里错了。 感谢所有的帮助。 这是ProfitHeader.h文件
#ifndef PROFITHEADER_H_
#define PROFITHEADER_H_
#include <string>
#include <map>
#include <vector>
using namespace std;
namespace kproblem{
typedef int Money;
typedef int Labor;
struct Resources{
Money liquidity;
Labor officeWork;
Labor programmingWork;
Resources(Money li, Labor of, Labor pro) : liquidity(li), officeWork(of), programmingWork(pro){}
//operator -=
Resources & operator -=( const Resources &rhs ){
liquidity -=rhs.liquidity;
officeWork -=rhs.officeWork;
programmingWork -=rhs.programmingWork;
return *this;
}
//operator< Used to make sure that key elements Match. will not modify (this)
bool operator<(const Resources & rhs) const{
if(this->liquidity < rhs.liquidity)
return true;
else if(this->liquidity > rhs.liquidity)
return false;
else if(this->officeWork < rhs.officeWork)
return true;
else if(this->officeWork > rhs.officeWork)
return false;
//this is the iff conditional
else if(this->programmingWork < rhs.programmingWork)
return true;
else
return false;
}
};
//Global Operator-. This will not modify (this).
Resources operator-( const Resources & lhs, const Resources & rhs ){
return Resources(lhs.liquidity - rhs.liquidity,
lhs.officeWork - rhs.officeWork, lhs.programmingWork - rhs.programmingWork);
}
//This is the Project Struct. It should contain the resources and data from the file.
struct Project{
string name;
Resources resources;
Money profit;
Project(string n, Resources re, Money p) : name(n), resources(re), profit(p) {}
};
//Definition of the ValueMap
typedef map<pair<Resources, vector<Project>::size_type>, pair<Money, bool>> ValueMap;
}
#endif
这是我的main.cpp
#include <iostream>
#include <fstream>
#include <sstream>
#include <exception>
#include "ProfitHeader.h"
using namespace std;
using namespace kproblem;
//The following was provided to us on the program
class IO_Exception : public runtime_error
{
public:
IO_Exception(const string & message) : runtime_error(message) { }
};
void readProjects(vector<Project> & projects, const string & fileName)
{
ifstream infile(fileName.c_str());
if (!infile)
throw IO_Exception("Could not open " + fileName);
string oneLine;
unsigned int lineNum = 0;
while (getline(infile, oneLine))
{
istringstream st(oneLine);
lineNum++;
string name;
Money liquidity;
Labor officeWork;
Labor programmingWork;
Money profit;
st >> name;
st >> liquidity;
st >> officeWork;
st >> programmingWork;
st >> profit;
if (st.fail())
{
cerr << "Skipping line number " << lineNum << ": "
<< oneLine << endl;
continue;
}
string junk;
if (st >> junk)
{
cerr << "Skipping line number " << lineNum << ": "
<< oneLine << endl;
continue;
}
projects.push_back(Project(name, Resources(liquidity, officeWork, programmingWork), profit));
}
if (!infile.eof())
throw IO_Exception("Error reading from " + fileName);
}
//Class Best Profit.
//This class will calculate the best possible profit we can get.
Money bestProfit(const vector<Project> & projects, Resources res, ValueMap & valMap,int n){
//initialize the best 2 possible solutions.
Money best1;
Money best2;
Money map; // the map where ou answers are stored
// First check if we are not at the end of the projects
if(n == 0){
return 0;
}
//now we are going to check the best project possible.
//Check the subinstance if it was solved.
if(valMap.find(make_pair(res, n-1)) != valMap.end()){
map = valMap.find(make_pair(res, n-1))->second.first;
return map;
}//check if the subinstance is solved. if it is return the value.
best1 = bestProfit(projects, res, valMap, n-1);//first best possible solution
//check the resources for the last project only. Fopr the second best possible solution.
if(res.liquidity >= projects.at(n-1).resources.liquidity
&& res.officeWork >= projects.at(n-1).resources.officeWork
&& res.programmingWork >= projects.at(n-1).resources.programmingWork){// feasability Check.
//all the above are requiered as it is necessary to check for all of them when doing the calculations.
best2 = bestProfit(projects, res - projects[n-1].resources, valMap, n-1) + projects[n-1].profit;
}
else{
best2 = 0;
}
//after the whole check compare the results and store the best possible result in the map.
if(best1 >= best2){
valMap.insert(make_pair(make_pair(res, n), make_pair(best1,false)));
return best1;
}
else{
valMap.insert(make_pair(make_pair(res, n), make_pair(best2,true)));
return best2;
}
}
//reportBestProfit. This will call Best profit and help us print the final results.
void reportBestProfit(vector<Project> projects, Resources resources){
ValueMap valueMap;
//Variables for the total resources used.
Money liq = 0;
Money ow = 0;
Money pw = 0;
int n = 1000; //number of projects, put here for fast testing
Money bestP = bestProfit(projects, resources, valueMap, n);
//Iterate the valuemap and print the best projects available to us.
cout << "Selected Projects -" << endl;
for(int i= 1; i <= 1000; i++){
//if(valueMap.find(make_pair(resources, i-1)) == valueMap.end()){
if(valueMap.find(make_pair(resources, i))->second.second == true){
//if(valueMap.find(make_pair(resources, i))->second.first != valueMap.find(make_pair(resources, i-1))->second.first){
//cout << valueMap.find(make_pair(resources, i))->second.first; //money
//cout <<" "<< valueMap.find(make_pair(resources, i))->second.second; //boolean
cout << " " << projects.at(i-1).name << " " << projects.at(i-1).resources.liquidity <<" ";//projects
cout << projects.at(i-1).resources.officeWork << " " << projects.at(i-1).resources.programmingWork;
cout << " " << projects.at(i-1).profit << endl;//profit
//}
}
}
cout << "Total Resources Used -" << endl;
//Print the resources consumed.
for(int i= 1; i <= 1000; i++){
if(valueMap.find(make_pair(resources, i))->second.second == true){
liq += projects.at(i-1).resources.liquidity;
ow += projects.at(i-1).resources.officeWork;
pw += projects.at(i-1).resources.programmingWork;
}
}
cout << " " << "Liquidity: " << liq <<endl;
cout << " " << "Office Work: " << ow <<endl;
cout << " " << "Programming Work: " << pw <<endl;
//Print the total Profit.
cout << "Profit: " << bestP << endl;
system("PAUSE");
}
int main()
{
vector<Project> projects;
try
{
readProjects(projects, "Proj5Data.txt");
}
catch (const IO_Exception & ex)
{
cerr << "IO error from: " << ex.what() << endl;
return 1;
}
//these values can be changed for different analysis on projects.
Money liquidity = 200;
Labor officeWork = 450;
Labor programmingWork = 1000;
cout << "Available resources - " << endl
<< " Liquidity: " << liquidity << endl
<< " Office Work: " << officeWork << endl
<< " Programming Work: " << programmingWork << endl;
reportBestProfit(projects, Resources(liquidity, officeWork, programmingWork));
return 0;
}
包含项目的项目文件可以在此暂时下载：
https://rapidshare.com/files/459861869/Proj5Data.txt
我的猜测是问题在于valmap查找，但我尝试了各种组合，但它根本不起作用。
最后这是我应该从中获得的最终打印输出：
但相反，我得到了所有其他结果，包括我需要的一些结果：
再次感谢你能够打我的头脑，并说，你好，你不应该再这样做了。）。
2 个答案:
答案 0 :(得分：0)
删除它会消除输出第二部分的前导数字
cout << valueMap.find(make_pair(resources, i))->second.first; //money
cout <<" "<< valueMap.find(make_pair(resources, i))->second.second; //boolean
cout << " "
此时打印的值未经过筛选和排序，这就是我认为您打印这些值的原因
但您没有打印"The total resources used -"部分
答案 1 :(得分：0)
好的，所以是的我有答案。现已完成（编辑后）
void reportBestProfit(vector<Project> projects, Resources resources){
ValueMap valueMap;
//Variables for the total resources used.
Money liq = 0;
Money ow = 0;
Money pw = 0;
vector<Project> result;
int n = 1000; //number of projects, put here for fast testing
Money bestP = bestProfit(projects, resources, valueMap, n);
//Iterate the valuemap and print the best projects available to us.
cout << "Selected Projects -" << endl;
// this loop just iterates through the values, it does not check the initial resources.
for(int i= 999; i > 0; i--){
//if(valueMap.find(make_pair(resources, i-1)) == valueMap.end()){
//check first If I still have resources available
if(resources.liquidity >=0 && resources.officeWork >= 0 && resources.programmingWork >= 0){
if(valueMap.find(make_pair(resources, i))->second.second == true){
//when I find the first true, I need to substract the resources of it from the base resources,
//to ask the question again.
resources.liquidity -= projects.at(i-1).resources.liquidity;
resources.officeWork -= projects.at(i-1).resources.officeWork;
resources.programmingWork -= projects.at(i-1).resources.programmingWork;
//Push the results into a vector for the printout
result.push_back(Project(projects.at(i-1).name,
Resources(projects.at(i-1).resources.liquidity,
projects.at(i-1).resources.officeWork,
projects.at(i-1).resources.programmingWork),
projects.at(i-1).profit));
//Also in one shot add together the resources used
liq += projects.at(i-1).resources.liquidity;
ow += projects.at(i-1).resources.officeWork;
pw += projects.at(i-1).resources.programmingWork;
}
}
}
//Print the saved vector in reverse order
for(int size = result.size(); size != 0; size--){
cout << " " << result.at(size -1).name;
cout << " " << result.at(size -1).resources.liquidity;
cout << " " << result.at(size -1).resources.officeWork;
cout << " " << result.at(size -1).resources.programmingWork;
cout << " " << result.at(size -1).profit << endl;
}
cout << "Total Resources Used -" << endl;
////Print the resources consumed.
cout << " " << "Liquidity: " << liq <<endl;
cout << " " << "Office Work: " << ow <<endl;
cout << " " << "Programming Work: " << pw <<endl;
//Print the total Profit.
cout << "Profit: " << bestP << endl;
system("PAUSE");
}
基本上我没有减少资源，所以我总是拥有过多的资源，但是一旦我做了那个中提琴！有用。谢谢你们看着它，我想我今天早上只需要灵感。
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