题目渊源：

        马踏棋盘问题（又称骑士周游问题或骑士漫游问题）是算法设计的经典问题之一。

题目要求：

        国际象棋的棋盘为8*8的方格棋盘，现将“马”放在任意指定的方格中，按照“马”走棋的规则将“马”进行移动。要求每个方格只能进入一次，最终使得“马”走遍棋盘64个方格。

        

#include <stdio.h>
#include <time.h>#define X 8
#define Y 8int chess[X][Y];//找到基于（x，y）位置的下一个可走的位置 
int nextxy(int *x,int *y,int count)
{switch(count){case 0:if(*x+2<=X-1 && *y-1>=0 && chess[*x+2][*y-1]==0){*y+=2;*y-=1;return 1;}break;case 1:if(*x+2<=X-1 && *y+1<=Y-1 && chess[*x+2][*y+1]==0 ){*x+=2;*y+=1;return 1;}break;case 2:if(*x+1<=X-1 && *y-2>=0 && chess[*x+1][*y-2]==0 ){*x=*x+1;*y=*y-2;return 1;}break;case 3:if(*x+1<=X-1 && *y+2<=Y-1 && chess[*x+1][*y+2]==0){*x = *x+1;*y= *y+2;return 1;}break;case 4:if(*x-2>=0  && *y-1>=0 && chess[*x-2][*y-1]==0){*x= *x-2;*y= *y+1;return 1;}break;case 5:if(*x-2>=0 && *y+1<=Y-1 && chess[*x-2][*y+1]==0 ){*x= *x-2;*y = *y+1;return 1;}break;case 6:if(*x-1>=0 && *y-2>=0 && chess[*x-1][*y-2]==0){*x = *x - 1;*y = *y - 2;return 1;}break;case 7:if(*x-1>=0 && *y+2<=Y-1 && chess[*x-1][*y+2]==0){*x = *x -1;*y = *y +2;return 1;}break;default:break;} return 0;
} void print()
{int i,j;for(i=0;i<X;i++){for(j=0;j<Y;j++){printf("%2d\t",chess[i][j]);}printf("\n");}printf("\n");
}//深度优先遍历棋盘
//（x，y）为位置坐标
//tag是标记变量
int TravelChessBoard(int x,int y,int tag)
{int x1= x,y1=y,count =0,flag =0;chess[x][y] = tag;if(x*Y == tag){//打印棋盘print();return 1; }//找到马的下一个可走的坐标（x1，y1）flag = nextxy(&x1,&y1,count);while(0==flag && count<7){count++;}while(flag){if(TravelChessBoard(x1,y1,tag+1)){return 1;}//出现意外，找到马的下一步可走坐标（x1，y1） x1=x;y1=y;count++;flag = nextxy(&x1,&y1,count);while(0==flag && count < 7){count++;flag = nextxy(&x1,&y1,count);}} if(0 == flag){chess[x][y] =0;} return 0;
} int main()
{int i,j;clock_t start,finish;start = clock();for(i=0;i<X;i++){for(j=0;j<Y;j++){chess[i][j]=0;}}if(TravelChessBoard(2,0,1)){printf("抱歉，马踏棋盘失败！\n");}finish = clock();printf("\n本次计算一共耗时:%f秒\n\n",(double)(finish - start)/CLOCKS_PER_SEC);return 0;
}