马踏棋盘，用1枚马走遍棋盘。我用一个二维数组记录模拟的整个路径，x为列，y为行，以顺时针的方式寻找下一格，算法比较简单，就通过递归和循环回溯即可，就是如果是8*8的数组，最坏可能执行8^(x*y)次，耗时长到怀疑人生。

#include<iostream>
#define X 5
#define Y 5

void ShowResult();
using namespace std;

int chess[Y][X]={
	0
};
int counter=0;

int Next(int* x,int* y,int where){

	switch(where){
		case 0:
			if(*x+1<X&&*y-2>=0&&chess[*y-2][*x+1]==0){
				*x+=1;
				*y-=2;
				return 1;
			}
			break;
		case 1:
			if(*x+2<X&&*y-1>=0&&chess[*y-1][*x+2]==0){
				*x+=2;
				*y-=1;
				return 1;
			}
			break;
		case 2:
			if(*x+2<X&&*y+1<Y&&chess[*y+1][*x+2]==0){
				*x+=2;
				*y+=1;
				return 1;
			}
			break;
		case 3:
			if(*x+1<X&&*y+2<Y&&chess[*y+2][*x+1]==0){
				*x+=1;
				*y+=2;
				return 1;
			}
			break;
		case 4:
			if(*x-1>=0&&*y+2<Y&&chess[*y+2][*x-1]==0){
				*x-=1;
				*y+=2;
				return 1;
			}
			break;
		case 5:
			if(*x-2>=0&&*y+1<Y&&chess[*y+1][*x-2]==0){
				*x-=2;
				*y+=1;
				return 1;
			}
			break;
		case 6:
			if(*x-2>=0&&*y-1>=0&&chess[*y-1][*x-2]==0){
				*x-=2;
				*y-=1;
				return 1;
			}
			break;
		case 7:
			if(*x-1>=0&&*y-2>=0&&chess[*y-2][*x-1]==0){
				*x-=1;
				*y-=2;
				return 1;
			}
			break;
	}
	return 0;
}

int Explore(int x,int y){
	int x1=x;
	int y1=y;
	int flag;
	int where=0;
	

	counter++;
	chess[y][x]=counter;
	

	if(counter==X*Y){
		return 1;
	}
	
	
	
	flag=Next(&x1,&y1,where);
	while(flag==0&&where<7){
		where++;
		flag=Next(&x1,&y1,where);
	}
	
	
	while(flag){
		if(Explore(x1,y1)==1){
			return 1;
		}
		else{
			x1=x;
			y1=y;
			where++;
			flag=Next(&x1,&y1,where);
			while(flag==0&&where<7){
				where++;
				flag=Next(&x1,&y1,where);
			}
		}
	}
	if(flag==0){
		chess[y][x]=0;
		counter--;
	}
	return 0;
}

void ShowResult(){
	
	for(int i=0;i<Y;i++){
		for(int j=0;j<X;j++){
			cout.width(4);
			cout<<chess[i][j]<<' ';
		}
		cout<<endl;
	}
	cout<<endl;
}

int main(){
	int start=clock();
	int result=Explore(2,1);
	int end=clock();
	if(result){
		ShowResult();		
	}
	else{
		cout<<"have no path！"<<endl; 
	}

	cout<<"spend time："<<(end-start)/CLOCKS_PER_SEC<<" s"<<endl;
	return 0;
}