C++实践参考:洗牌(范型程序设计)

【项目2:洗牌】
  在扑克牌游戏中,每次游戏开始都要求把54张牌重新排列一下,称为洗牌。试编写程序将一副扑克牌(用54个整数1~54表示)随机洗好后,顺序输出54张牌的情况。
  参考界面:
C++实践参考:洗牌(范型程序设计)_第1张图片

参考解答(共4种,可作为程序阅读,品味用STL解决问题的方法,必要时,请查找相关手册)
  解法1:初始化一个 vector,顺序加入所有牌,即整数1~54。然后从容器中随机抽取一个加到另一个vector中,这个过程一共执行54次。

#include <ctime>
#include <vector>
#include <list>
#include <iostream>
#include <iterator>
#include <cstdlib>
using namespace std;

using namespace std;
typedef vector<int> IntVector;
typedef unsigned int VIndex;
void vectorShuffle(IntVector &unshuffled,IntVector &shuffled)
{
    VIndex p,size=unshuffled.size();
    while(size)
    {
        p=rand()%size--;
        shuffled.push_back(unshuffled[p]);
        unshuffled.erase(unshuffled.begin()+p);
    }
}

int main()
{
    ostream_iterator<int> os(cout," ");
    srand(time(NULL));
    IntVector c,sc;
    for(VIndex i=1; i<=54; i++)
    {
        c.push_back(i);
    }
    cout<<"Before Shuffle"<<endl;
    copy(c.begin(),c.end(),os);
    cout<<endl;
    vectorShuffle(c,sc);
    cout<<"\nAfter Shuffled"<<endl;
    copy(sc.begin(),sc.end(),os);
    cout<<endl<<endl;
    return 0;
}

  解法2:相同思路,用list

#include <ctime>
#include <vector>
#include <list>
#include <iostream>
#include <iterator>
#include <cstdlib>
using namespace std;

typedef list<int> IntList;
typedef unsigned int VIndex;

void listShuffle(IntList &unshuffled,IntList &shuffled)
{
    VIndex  p, size=unshuffled.size();
    IntList::iterator iter;
    while(size)
    {
        p=rand()%size--;
        iter=unshuffled.begin();
        while(p!=0)
        {
            iter++;
            p--;
        }
        shuffled.push_back(*iter);
        unshuffled.erase(iter);
    }
}
int main()
{
    ostream_iterator<int> os(cout," ");
    srand(time(NULL));
    IntList cl,scl;
    for(VIndex i=1; i<=54; i++)
    {
        cl.push_back(i);
    }
    cout<<"Before Shuffle"<<endl;
    copy(cl.begin(),cl.end(),os);
    cout<<endl;
    listShuffle(cl,scl);
    cout<<"\nAfter Shuffled"<<endl;
    copy(scl.begin(),scl.end(),os);
    cout<<endl<<endl;
    return 0;
}

  解法3:随机交换两个位置的元素来洗牌。函数中time是要执行交换的次数,如果是54张牌的话,交换次数大于27的话就已经表现出很随机的排列了。

#include <ctime>
#include <vector>
#include <iterator>
#include <algorithm>
#include <iostream>
using namespace std;
typedef vector<int> IntVector;
void SwapShuffle(IntVector &datas, int time)
{
    unsigned size=datas.size(),p1,p2;
    while(time--)
    {
        p1=rand()%size;
        p2=rand()%size;
        swap(datas[p1],datas[p2]);
    }
}
int main()
{
    ostream_iterator <int>  os(cout," ");
    srand(time(NULL));
    vector <int> poker;
    for(int i=1; i<=54; i++)
    {
        poker.push_back(i);
    }
    cout<<"Before Shuffle"<<endl;
    copy(poker.begin(),poker.end(),os);
    cout<<endl;
    SwapShuffle(poker,100);
    cout<<"\nAfter Shuffled"<<endl;
    copy(poker.begin(),poker.end(),os);
    cout<<endl<<endl;
    return 0;
}

  解法4:采用STL的 random_shuffle 算法

#include <ctime>
#include <vector>
#include <iterator>
#include <algorithm>
#include <iostream>
using namespace std;
int main()
{
    ostream_iterator <int>  os(cout," ");
    srand(time(NULL));  // 洗牌前要先初始化随机数种子
    vector <int> poker;
    for(int i=1; i<=54; i++)
    {
        poker.push_back(i);
    }
    cout<<"Before Shuffle"<<endl;
    copy(poker.begin(),poker.end(),os);
    cout<<endl;
    random_shuffle(poker.begin(),poker.end());
    cout<<"\nAfter Shuffled"<<endl;
    copy(poker.begin(),poker.end(),os);
    cout<<endl<<endl;
    return 0;
}

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