此代码试图解决 4 皇后问题,将 4 个皇后放在一个 4*4 的棋盘上,但其中任何一个都无法捕获对方
#include <iostream>
using namespace std;
int Place(int Chess[][4], int collumn, int i);
bool Check(int Chess[][4], int collumn, int i);
int findrow(int Chess[][4], int collumn);
const int size = 3;
int main()
{
int Chess[4][4];
int collumn;
int i = 0;
collumn = 0;
for(int s = 0; s < 4; s++)
{
for(int j = 0; j < 4; j ++)
{
Chess[s][j] = 0;
}
}
//Chess[0][0] = 1;
//Chess[3][3] = 1;
//if(Check(Chess, 3, 3) == false)
Place(Chess, collumn, i);
for(int z = 0; z < 4; z++)
{
for(int a = 0; a < 4; a++)
{
if(Chess[z][a] == 1)
cout<<"Row: "<<z<<"Collumn: "<<a<<"."<<endl;
}
cout<<endl;
}
system("pause");
return 0;
}
int Place(int Chess[][4], int collumn, int i)
{
if(collumn > size)
return 0;
while(i <= size)
{
if(Check(Chess, collumn, i) == true)
{
//cout<<"hi"<<endl;
Chess[collumn][i] = 1;
return(Place(Chess, (collumn + 1), i));
}
i ++;
}
if(i>= size)
{
//cout<<"hilo"<<endl;
return Place(Chess, collumn-1, findrow(Chess, collumn-1));
}
}
bool Check(int Chess[][4], int collumn, int i)//checks to see if it can be captured
{// very inneficitnt
int x = collumn;// this is so we can now work in terms of x and y
int y = i;
bool found = true;
// checks all the diagonal captures
if(Chess[x -1 ][y -1]== 1&& x>=1 && y >=1 )
found = false;
if(Chess[x -2 ][y - 2]== 1&& x>=2 && y>=2 )
found = false;
if(Chess[x - 3][y - 3]== 1 && x>=3 && y>=3 )
found = false;
if(Chess[x + 1][y - 1] == 1&& x<=2 && y>=1 )
found = false;
if(Chess[x + 2][y -2] == 1&& x<=1 && y>=2)
found = false;
if(Chess[x + 3][y - 3] == 1 && x<=0 && y>=3)
found = false;
if(Chess[x + 1][y + 1] == 1 && x<=2 && y<=2)
found = false;
if(Chess[x + 2][y + 2] == 1&& x<=1 && y<=1)
found = false;
if(Chess[x + 3][y + 3] == 1 && x<=0 && y<=0 )
found = false;
if(Chess[x -1 ][y + 1]== 1 && x>=1 && y<=2 )
found = false;
if(Chess[x - 2][y + 2] == 1&& x>=2 && y<=1 )
found = false;
if(Chess[x - 3][y + 3] == 1&& x>=3 && y<=0)
found = false;
//checks all the horizontal captures. We don't need to check for vertical captures
if(Chess[x + 1][y] == 1 && x<=2)
found = false;
if(Chess[x + 2][y] == 1&& x<=1 )
found = false;
if(Chess[x+3][y] == 1 && x<=0)
found = false;
if(Chess[x -1 ][y] == 1&& x>=1)
found = false;
if(Chess[x-2][y] == 1&& x>=2 )
found = false;
if(Chess[x-3][y] == 1 && x>=3)
found = false;
if(found == false)
return false;
if(found == true)
return true;
}
int findrow(int Chess[][4], int collumn)
{
for(int z = 0; z < 4; z++)
{
if(Chess[collumn][z] == 1)
{
Chess[collumn][z] = 0;
return z;
}
}
}
最佳答案
我首先看到的是可能的越界访问:
if(Chess[x -1 ][y -1]== 1&& x>=1 && y >=1 )
如果 x
的值为 0
怎么办?您正在访问超出范围的 Chess[-1][y]。您的 if
语句不会停止此操作,即使在 x>=1
条件下也是如此。
if
将首先测试 Chess[x-1][y-1]==1
条件。如果您不希望发生这种情况,请将 x>=1
的测试放在 Chess[x-1][y-1]==1
之前。
但即便如此,整段代码看起来也很可疑。如果有更多的越界访问,我不会感到惊讶。
关于c++ - 我的 C++ 程序运行,但说 "8queens.exe has stopped working",我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/23587535/