以下是我写的代码..我必须为nXn编写它，但为了方便我试图测试它为5X5。它没有显示我的输出......任何人都可以告诉我以下代码有什么问题：
{
#include <iostream>
#include <iomanip>
using namespace std;
void knight ( int startx, int starty, int n, int p[][5], int used [][5], int &count);
int main ( )
{
const int n = 5; // no. of cloumns and rows
int startx = 0;
int starty = 0;
int p[5][5];
int used[5][5];
int count = 1;
int i= 0;
int j = 0;
//initializing the array
for ( i = 0; i < 5; i++)
{
for ( j = 0; j < 5; j++)
{
p[i][j] = 0;
used [i][j] = 0;
}
}
//outputting the initialized array..
i=0;
while ( i< 5)
{
for ( j = 0; j < 5; j++)
{
cout << setw(3) << p[i][j];
}
i++;
cout << endl;
}
knight (startx,starty,n,p,used,count);
return 0;
}
void knight ( int x, int y, int n, int p[][5], int used [][5], int &count)
{
int i = 0;
//knight (x,y,n,p,used,count)
for ( i = 0; i < n*n; i++)
{
if ( used [x][y] == 0 )
{
used[x][y] = 1; // mark it used;
p[x][y] += count; //inserting step no. into the solution
//go for the next possible steps;
//move 1
//2 squares up and 1 to the left
if (x-1 < 0 && y+2 < n && p[x-1][y+2] == 0)
{
used[x-1][y+2] = 1;
p[x-1][y+2] += count;
knight (x-1,y+2,n,p,used,count);
used[x-1][y+2] = 0;
}
//move 2
//2 squares up and 1 to the right
if ( x+1 < n && y+2 < n && p[x+1][y+2] == 0 )
{
used[x+1][y+2] = 1;
p[x+1][y+2] += count;
knight (x+1,y+2,n,p,used,count);
used[x+1][y+2] = 0;
}
//move 3
//1 square up and 2 to the right
if ( x+2 < n && y+1 < n && p[x+2][y+1] == 0 )
{
used[x+2][y+1] = 1;
p[x+2][y+1] += count;
knight (x+2,y+1,n,p,used,count);
used[x+2][y+1] = 0;
}
//move 4
//1 square down and 2 to the right
if ( x+2 < n && y-1 < n && p[x+2][y-1] == 0 )
{
used[x+2][y-1] = 1;
p[x+2][y-1] += count;
knight (x+2,y-1,n,p,used,count);
used[x+2][y-1] = 0;
}
//move 5
//2 squares down and 1 to the right
if ( x+1 < n && y-2 < n && p[x+1][y-2] == 0 )
{
used[x+1][y-2] = 1;
p[x+1][y-2] += count;
knight (x+1,y-2,n,p,used,count);
used[x+1][y-2] = 0;
}
//move 6
//2 squares down and 1 to the left
if ( x-1 < n && y-2 < n && p[x-1][y-2] == 0 )
{
used[x-1][y-2] = 1;
p[x-1][y-2] += count;
knight (x-1,y-2,n,p,used,count);
used[x-1][y-2] = 0;
}
//move 7
//1 square down and 2 to the left
if ( x-2 < n && y-1 < n && p[x-2][y-1] == 0 )
{
used[x-2][y-1] = 1;
p[x-2][y-1] += count;
knight (x-2,y-1,n,p,used,count);
used[x-2][y-1] = 0;
}
//move 8
//one square up and 2 to the left
if ( x-2 < n && y+1< n && p[x-2][y+1] == 0 )
{
used[x-2][y+1] = 1;
p[x-2][y+1] += count;
knight (x-2,y+1,n,p,used,count);
used[x-2][y+1] = 0;
}
}
}
if ( x == n-1 && y == n-1)
{
while ( i != n)
{
for ( int j = 0; j < n; j++)
cout << setw(3) << p[i][j];
i++;
}
}
}
谢谢！
答案 0 :(得分：1)
首先，一些编译器很挑剔，如果函数声明和定义中的变量名称不同，它们会给你一个警告。
void knight ( int startx, int starty, int n, int p[][5], int used [][5], int &count);
[...]
void knight ( int x, int y, int n, int p[][5], int used [][5], int &count) { ... }
其次，如果您尝试将int指针传递给骑士（参数count），那么您将使用int *count。如果您要在knight中使用递归，那么最好将count声明为全局，或者（如此处所示）如果count参数仅对该特定有效调用树，您可能不想使用指针，只需为每个调用提供自己的count。
第三，文件顶部的流氓{是什么？你的意思是把它放在那里吗？
运行程序时程序会执行什么操作？您是否尝试插入在代码的各个部分中打印出来的调试消息，以便您可以查看错误或无限循环的发生位置？
答案 1 :(得分：1)
这些都是代码中不同问题的提示：
在if ( x == n-1 && y == n-1)执行的程序中，i有什么价值？它是如何获得这个价值的？
这个循环for ( i = 0; i < n*n; i++) 实际有什么影响？
当你有y-2 < n这样的表达式时，你真正想要的是什么？ y-2
答案 2 :(得分：1)
我使用以下代码得到了答案：
#include <iostream>
#include <iomanip>
using namespace std;
bool move(int *p[] ,int x, int y,int n);
void knights (int *p[],int x,int y,int n, int count, int &pmt);
void output(int *p[],int n);
int main(char argc, char *argv[])
{
int count = 1;
int pmt = 0;
int n; //for size of board
int x,y; // starting pos
int **p; // to hold no. of combinations
if ( argc != 4)
{
cout << "Very few arguments. Please try again.";
cout << endl;
return 0;
}
n = atoi(argv[1]);
if( argv[1] <= 0 )
{
cout << " Invalid board size. ";
return 0;
}
x = atoi(argv[2]);
y = atoi(argv[3]);
cout << "board size: " << n << ", "<< n << endl;
cout << "starting pos: " << x << ", " << y << endl;
//dynamic allocation of arrays to hold permutation
p = new int *[n];
for (int i = 0; i < n; i++)
p[i] = new int [n];
//initializing board
int i, j;
for (i=0; i<n; i++)
{
for (j=0; j<n; j++)
{
p[i][j] = 0;
}
}
output(p,n); // check output format
knights(p,x,y,n,count,pmt);
cout << "no. perm " << pmt << endl;
return 0;
}
void output(int *p[],int n)
{
int i = 0;
int j;
while ( i != n)
{
for ( j=0; j<n; j++)
{
cout << setw(10) << p[i][j];
}
cout << endl;
i++;
}
}
bool move(int *p[],int x, int y,int n)
{
if (x < 0 || x >= n)
{
return false;
}
if ( y < 0 || y >= n)
{
return false;
}
if( p[x][y] != 0 )
{
return false;
}
return true;
}
void knights (int *p[], int x,int y,int n ,int count , int &pmt)
{
if(!move(p,x,y,n))
{
return;
}
if (count == n*n)
{
cout << "New solution found: " << endl <<endl;
p[x][y]=count;
output(p,n);
p[x][y]= 0;
pmt++;
}
if ( p[x][y] == 0 )
{
p[x][y] = count;
//move 1
knights (p, x-1, y-2, n, count+1, pmt);
//move 2
knights (p, x+1, y-2, n, count+1, pmt);
//move 3
knights (p, x+2, y-1, n, count+1, pmt);
//move 4
knights (p, x+2, y+1, n, count+1, pmt);
//move 5
knights (p, x+1, y+2, n, count+1, pmt);
//move 6
knights (p, x-1, y+2, n, count+1, pmt);
//move 7
knights (p, x-2, y+1, n, count+1, pmt);
//move 8
knights (p, x-2, y-1, n, count+1, pmt);
p[x][y] = 0; //setting square back to zero
}
}