马踏棋盘问题

问题描述：将马随机放在国际象棋的Board[0~7][0~7]的某个方格中，马按走棋规则进行移动。走遍棋盘上全部64个方格并且不能重复。求出马的行走路线，并按求出的行走               路线，将数字1，2，…，64依次填入一个8×8的方阵，输出之。

  1 #include<stdio.h>
  2 #include<time.h>
  3 
  4 #define CHESS_SIZE 5
  5 
  6 int chess[CHESS_SIZE][CHESS_SIZE];
  7 
  8 int getNextPosition(int* x, int* y, int counter){
  9     switch(counter)
 10     {
 11         case 1:
 12             if( *x - 2 >= 0 && *y - 1 >= 0 && chess[*x-2][*y-1] == 0 ){
 13                 *x -= 2;
 14                 *y -= 1;
 15                 return 1;
 16             }
 17             break;
 18         case 2:
 19             if( *x - 2 >= 0 && *y + 1 < CHESS_SIZE && chess[*x-2][*y+1] == 0 ){
 20                 *x -= 2;
 21                 *y += 1;
 22                 return 1;
 23             }
 24             break;
 25         case 3:
 26             if( *x - 1 >= 0 && *y + 2 < CHESS_SIZE && chess[*x-1][*y+2] == 0 ){
 27                 *x -= 1;
 28                 *y += 2;
 29                 return 1;
 30             }
 31             break;
 32         case 4:
 33             if( *x + 1 < CHESS_SIZE && *y + 2 < CHESS_SIZE && chess[*x+1][*y+2] == 0 ){
 34                 *x += 1;
 35                 *y += 2;
 36                 return 1;
 37             }
 38             break;
 39         case 5:
 40             if( *x + 2 < CHESS_SIZE && *y + 1 < CHESS_SIZE && chess[*x+2][*y+1] == 0 ){
 41                 *x += 2;
 42                 *y += 1;
 43                 return 1;
 44             }
 45             break;
 46         case 6:
 47             if( *x + 2 < CHESS_SIZE && *y - 1 >= 0 && chess[*x+2][*y-1] == 0 ){
 48                 *x += 2;
 49                 *y -= 1;
 50                 return 1;
 51             }
 52             break;
 53         case 7:
 54             if( *x + 1 < CHESS_SIZE && *y - 2 >= 0 && chess[*x+1][*y-2] == 0 ){
 55                 *x += 1;
 56                 *y -= 2;
 57                 return 1;
 58             }
 59             break;
 60         case 8:
 61             if( *x - 1 >= 0 && *y - 2 >= 0 && chess[*x-1][*y-2] == 0 ){
 62                 *x -= 1;
 63                 *y -= 2;
 64                 return 1;
 65             }
 66             break;
 67         default:
 68             break;
 69     }
 70     return 0;
 71 }
 72 int horseStepBoard(int x, int y, int step){
 73     int flag = 0, counter = 1;
 74     int x1 = x;
 75     int y1 = y;
 76     chess[x][y] = step;            //已经走到step
 77     if( CHESS_SIZE * CHESS_SIZE == step )
 78         return 1;            //遍历结束,任务完成
 79                         //还没有结束的话，找step的下一步在哪里
 80     flag = getNextPosition(&x1,&y1,counter);    
 81     while( flag == 0 && counter <= 8 ){
 82         counter++;
 83         flag = getNextPosition(&x1, &y1, counter);
 84     }                    //如果找到next，则x1, y1的值会被改变。若没有找到，不执行下面的while，
 85                         //说明x和y的值不合适，即周围至多的8个位置都不合适，将chess[i][j]还原
 86                         //为0，跳出本层递归至代码90行，并且携带返回值0；在上层递归中找(x,y)
 87                         //的nextPosition
 88     
 89     while( flag ){                //找到了下一步
 90         if( horseStepBoard(x1, y1, step+1) == 1 )
 91             return 1;
 92         x1 = x;
 93         y1 = y;
 94         flag = getNextPosition(&x1, &y1, counter++);
 95         while( flag == 0 && counter <= 8 ){
 96             counter++;
 97             flag = getNextPosition(&x1, &y1, counter);
 98         }
 99     }
100 
101     if( flag == 0 ){
102         chess[x][y] = 0;
103         return 0;
104     }
105 }
106 void PrintChess(){
107     int i, j;
108     for(i = 0; i < CHESS_SIZE; i++){
109         for(j = 0; j < CHESS_SIZE; j++)
110             printf("%5d", chess[i][j]);
111         printf("\n\n");
112     }
113 }
114 int main(){
115     clock_t start, finish;
116     start = clock();
117     horseStepBoard(2,0,1);
118     PrintChess();
119     finish = clock();
120     printf("running time is %fs\n", (double)(finish - start)/(CLOCKS_PER_SEC));
121     return 0;
122 }

 

posted @ 2016-03-11 15:46  robin_X  阅读(487)  评论(0)  编辑  收藏  举报