一道《C语言版数据结构》的题目！帮帮忙啊！

#include
#include
#include

const int N = 8;

int w = 0;
int way1[8] = {-2, -1, 1, 2, 2, 1, -1, -2};
int way2[8] = { 1, 2, 2, 1, -1, -2, -2, -1};
int ch[N*N] = { 0 };
int a[N*N+1][3] = { 0 };
int dir[N][N][8];
int st = 1;
char c = 'y';
int weight[N][N];

void caculate();
void dirctions();
void print();
int check(int i, int j);

void caculate() { //计算各点的权值
int i, j, k;
for(i = 1; i <= N; i++)
for(j = 1; j <= N; j++)
for(k = 0; k < N; k++) {
int x, y;
x = i + way1[k];
y = j + way2[k];
if ( x >= 1 && x <= N && y >= 1 && y<=N )
weight[i-1][j-1]++;
}
}

int check(int i,int j) { //检查(i,j)是否在棋盘内
if( i < 1 || i > 8 || j < 1 || j > 8 )
return 0;
return 1;
}

void directions() { //求出各点的最佳方向序列，即优先向权值小的方向
int i, j, k, m, n1, n2, x1, y1, x2, y2, way_1, way_2;
for(i = 0; i < N; i++)
for(j = 0; j < N; j++) {
for(k = 0; k < 8; k++)
dir[i][j][k] = k;
for(k = 0; k < 8; k++) {
for(m = k + 1; m < 8; m++) { //对每个方向考察看有没有更好的
way_1 = dir[i][j][k];
x1 = i + way1[way_1];
y1 = j + way2[way_1];
way_2 = dir[i][j][m];
x2 = i + way1[way_2];
y2 = j + way2[way_2];
n1 = check(x1+1,y1+1);
n2 = check(x2+1,y2+1);
if( ( n1==0 && n2 ) || //k方向不可达到，而m方向可达到
( n1 && n2 && weight[x1][y1] > weight[x2][y2] )//都可达到但m方向权值小
) {
dir[i][j][k] = way_2;
dir[i][j][m] = way_1; //交换两个方向值
}
}
}
}
}

void print() {
int x, y;
cout<<"\n---------"<<++w<<"answer---------\n";
for(x = 1; x < N + 1; x++) {
cout< for(y = 1; y < N + 1; y++)
cout< cout< }
cout<<"\nPress n to quit ,press any other key to continue.\n";
c = getchar(); //询问是否继续输出结果
}

void main() {
int x, y, way, way0;
caculate();
directions();
cout<<"Please enter the row and column of the starting point.\n";
cin>>a[1][0]>>a[1][1]; //输入行数和列数
getchar(); //接收回车符
x = a[1][0], y = a[1][1];
ch[(x-1)*N+y-1] = 1; //在ch数组中对相应点赋值
while(1) {
if(a[1][2] >= 8) //出发点的八个方向都已走过，表示所有的方法均已找出
break;
if(a[st][2] >= 8) { //此点的八个方向都已走过，应该退回到上一次走的点
x = a[st][0];
y = a[st][1];
ch[(x-1)*N+y-1] = 0; //将这一点被走过的痕迹抹去
a[st][0] = a[st][1] = a[st][2] = 0;
a[st-1][2]++; //使上一次走的点走的方向发生变化
st--; //步数减一
}
else { //此点的八个方向未全走过，应走此方向
way0 = a[st][2];
a[st][2]++; //确定下次应走的方向
x = a[st][0];
y = a[st][1];
way = dir[x-1][y-1][way0];
x = a[st][0] + way1[way];
y = a[st][1] + way2[way]; //确定按这次的方向走应走到的x,y坐标
if(x<1||y<1||x>N||y>N||ch[(x-1)*N+y-1]!=0) //此点不满足要求
continue;
ch[(x-1)*N+y-1] = ++st; //走到这一点
a[st][0] = x;
a[st][1] = y;
a[st][2] = 0; //标记这一步
if(st == N*N) { //步数已满
print(); //输出结果
if(c == 'n')
break;
ch[(x-1)*N+y-1] = 0;
a[st][0] = a[st][1] = a[st][2] = 0;
a[st-1][2]++;
st--; //退回前一步
}
}
}
}

这样一个小程序写那么麻烦干嘛……简单的回溯法搞定

#include
const int m=8,n=8;
const int dir[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};
int start_x,start_y;
int flag[n][m];
int stack[m*n][2];
int done;

void find(int depth)
{
if (done) return;
if (depth==m*n)
{
int i;
for (i=0;iprintf("%d %d\n",stack[i][0],stack[i][1]);
done=1;
}
else
{
int i;
int x,y;
x=stack[depth-1][0];
y=stack[depth-1][1];
for (i=0;!done && i<8;i++)
if (x+dir[i][0]>=0 && x+dir[i][0]=0 && y+dir[i][1]if (flag[x+dir[i][0]][y+dir[i][1]])
{
flag[x+dir[i][0]][y+dir[i][1]]=0;
stack[depth][0]=x+dir[i][0];
stack[depth][1]=y+dir[i][1];
find(depth+1);
flag[x+dir[i][0]][y+dir[i][1]]=1;
}
}
}

main()
{
int i,j;
for (i=0;ifor (j=0;jscanf("%d%d",&start_x,&start_y);
flag[start_x][start_y]=0;
stack[0][0]=start_x;
stack[0][1]=start_y;
done=0;
find(1);
}