皇后问题

  1 #include"iostream"
  2 #include"windows.h"
  3 using namespace std;
  4 
  5 class Piece{
  6 private:
  7     int *p;
  8     int n;
  9 public:
 10     static int count;
 11     static char **map;
 12     
 13     //创建棋子
 14     Piece(int n);
 15     
 16     //创建棋盘
 17     void createMap(char bg,char piece);
 18     
 19     //打印棋盘
 20     void show();
 21     
 22     //判断棋子是否可放
 23     bool judge(int x);
 24     
 25     //移动棋子
 26     void move(int x);
 27 };
 28 
 29 int Piece::count = 0;
 30 char **Piece::map;
 31 int main(){
 32     char bg = '+';
 33     char pie = '#';
 34     int n = 8;
 35     Piece p(n);
 36     p.createMap(bg,pie);
 37     p.move(0);
 38     return 0;
 39 }
 40 
 41 Piece::Piece(int n){
 42     this->n = n;
 43     p = new int[n];
 44     for(int i = 0;i < n;i++){
 45         p[i] = -1;
 46     }
 47 }
 48 void Piece::createMap(char bg,char piece){
 49     map = new char*[n];
 50     for(int i = 0;i < n;i++){
 51         map[i] = new char[n];
 52     }
 53     for(i = 0;i < n;i++){
 54         for(int j = 0;j < n;j++){    
 55             map[i][j] = bg;
 56         }
 57     }
 58 }
 59 
 60 void Piece::show(){
 61     for(int i = 0;i < n;i++){
 62         for(int j = 0;j < n;j++){
 63             if(p[j] == i){
 64                 cout<<"O"<<" ";
 65             }
 66             else{
 67                 cout<<map[i][j]<<" ";
 68             }
 69         }
 70         cout<<endl;
 71     }
 72 }
 73 
 74 void Piece::move(int x){
 75     if(x<n){            //棋子在棋盘内
 76         for(int i = 0;i < n;i++){
 77             p[x] = i;
 78             if(judge(x)){            //棋子可以放
 79                 if(x == n-1){
 80                     cout<<"-------"<<++count<<"-------"<<endl;
 81                     this->show();
 82                 }
 83                 move(x + 1);        //放下一个棋子
 84             }
 85             //棋子不能放，放下一个
 86         }
 87         //都不能放，不放，重新放上一个
 88         p[x] = -1;
 89     }
 90 }
 91 bool Piece::judge(int x){            //第x列
 92     int x1 = x + p[x];        //反斜线
 93     int x2 = x - p[x];        //正斜线
 94     for(int i = 0;i < n;i++){
 95         if(p[i]!=-1&&i!=x&&p[x] == p[i]){        //行
 96             return false;
 97         }
 98         if(x2+i!=x&&x2 + i>=0&&x2 + i<n&&p[x2 + i] != -1&&x2 + i - x == p[x2 + i] - p[x]){        //正斜线
 99             return false;
100         }
101         if(x1 - i!=x&&x1 - i>=0&&x1 - i<n&&p[x1 - i] != -1&&x1 - i - x == p[x] - p[x1 - i]){        //反斜线
102             return false;
103         }
104     }
105     return true;
106 }
BY oleolema