C++贪心算法实现活动安排问题(实例代码)

贪心算法

贪心算法（又称贪婪算法）是指，在对问题求解时，总是做出在当前看来是最好的选择。也就是说，不从整体最优上加以考虑，他所做出的是在某种意义上的局部最优解。

贪心算法不是对所有问题都能得到整体最优解，关键是贪心策略的选择，选择的贪心策略必须具备无后效性，即某个状态以前的过程不会影响以后的状态，只与当前状态有关。

具体代码如下所示：

#include <cstdio>
#include <iostream>
#include <ctime>
#include <windows.h>
#include <algorithm>
#include <fstream>
using namespace std;
struct activity
{
  int no;
  int start;
  int finish;
};
bool cmp(const activity &x, const activity &y)
{
  return x.finish<y.finish;//从小到大排<，若要从大到小排则>
}
int greedySelector(int m,int solution[],struct activity activity[]){
  int number = 1;
  solution[0] = 1;
  int i,j = 0,counter = 1;
  for(i = 1;i < m ;i++)
  {
    if(activity[i].start >=activity[j].finish)
    {
      solution[i] = 1;
      j = i;
      counter++;
    }
    else
      solution[i] = 0;
  }
  cout << "The amount of activities is:"<<counter<<endl;
  cout << "The solution is:";
  for(i = 0 ;i < m ;i++)
  {
    if (solution[i] == 1)
    {
      cout << activity[i].no <<" ";
    }
  }
  return counter;
}
int main(void)
{
  LARGE_INTEGER nFreq;
  LARGE_INTEGER nBeginTime;
  LARGE_INTEGER nEndTime;
  ofstream fout;
  srand((unsigned int)time(NULL));
  int m,i,j,t;
  double cost;
  cout << "Please enter the number of times you want to run the program:";
  cin >> t;
  fout.open("activity.txt",ios::app);
  if(!fout){
    cerr<<"Can not open file 'activity.txt' "<<endl;
    return -1;
  }
  fout.setf(ios_base::fixed,ios_base::floatfield);    //防止输出的数字使用科学计数法
  for (j = 0;j < t;j++)
  {
    cout << "——————————————————The "<< j + 1 << "th test —————————————————"<<endl;
    m = 1 + rand()%100000;
    fout<<m<<",";
    int solution[m];
    activity activity[m];
    for( i = 0;i < m;i++)
    {
      activity[i].no = i+1;
      activity[i].start = 1 + rand()%1000;
      while(1)
      {
        activity[i].finish = 1 + rand()%10000;
        if(activity[i].finish > activity[i].start) break;
      }
    }
    QueryPerformanceFrequency(&nFreq);
    QueryPerformanceCounter(&nBeginTime);
    sort(activity,activity+m,cmp);
    greedySelector(m,solution,activity);
    QueryPerformanceCounter(&nEndTime);
    cost=(double)(nEndTime.QuadPart - nBeginTime.QuadPart) / (double)nFreq.QuadPart;
    fout << cost << endl;
    cout << "\nThe running time is:" << cost << " s" << endl;
  }
  fout.close();
  cout << endl << endl;
  cout << "Success!" << endl;
  return 0;
}

总结

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