骑士走棋盘

  1 #include <iostream>
  2 #include <cstdio>
  3 
  4 using namespace std;
  5 
  6 int pos[8][8] = {0};
  7 
  8 int travel(int, int);
  9 
 10 int main()
 11 {
 12     int i, j, startX, startY;
 13     printf("Please input a starting point: ");
 14     scanf("%d%d", &startX, &startY);
 15     if(travel(startX, startY)) {
 16         printf("Travel finished\n");
 17     }else {
 18         printf("Travel failed\n");
 19     }
 20     for(i=0; i<8; i++) {
 21         for(j=0; j<8; j++) {
 22             printf("%2d ", pos[i][j]);
 23         }
 24         printf("\n");
 25     }
 26     return 0;
 27 }
 28 
 29 int travel(int x, int y) {
 30     int i, j, k, l, m;
 31     int tmpX, tmpY;
 32     int count, min, tmp;
 33 
 34     //骑士可走的八个方向(顺时针)
 35     int ktmoveX[8] = {1, 2, 2, 1, -1, -2, -2, -1};
 36     int ktmoveY[8] = {-2, -1, 1, 2, 2, 1, -1, -2};
 37 
 38     //下一步坐标
 39     int nextX[8] = {0};
 40     int nextY[8] = {0};
 41 
 42     //记录每个方向的出路的个数
 43     int exists[8] = {0};
 44 
 45     //起始用1标记位置
 46     i = x;
 47     j = y;
 48     pos[i][j] = 1;
 49 
 50     //遍历棋盘
 51     for(m=2; m<=64; m++) {
 52         //初始化八个方向出口个数
 53         for(l=0; l<8; l++) {
 54             exists[l] = 0;
 55         }
 56         l = 0; //计算可走方向
 57 
 58         //试探八个方向
 59         for(k=0; k<8; k++) {
 60             tmpX = i + ktmoveX[k];
 61             tmpY = j + ktmoveY[k];
 62             //边界 跳过
 63             if(tmpX<0 || tmpY<0 || tmpX>7 || tmpY>7) {
 64                 continue;
 65             }
 66             //可走 记录
 67             if(pos[tmpX][tmpY] == 0) {
 68                 nextX[l] = tmpX;
 69                 nextY[l] = tmpY;
 70                 l++;    //可走方向加1
 71             }
 72         }
 73         count = l;
 74         //无路可走 返回
 75         if(count == 0) {
 76             return 0;
 77         //一个方向可走 标记
 78         }else if(count == 1) {
 79             min = 0;
 80         //找出下个位置出路个数
 81         }else {
 82             for(l=0; l<count; l++) {
 83                 for(k=0; k<8; k++) {
 84                     tmpX = nextX[l] + ktmoveX[k];
 85                     tmpY = nextY[l] + ktmoveY[k];
 86                     if(tmpX<0 || tmpY<0 || tmpX>7 || tmpY>7) {
 87                         continue;
 88                     }
 89                     if(pos[tmpX][tmpY] == 0) {
 90                         exists[l]++;
 91                     }
 92                 }
 93             }
 94             //找出下个位置出路最少的方向
 95             min = 0;
 96             tmp = exists[0];
 97             for(l=0; l<count; l++) {
 98                 if(exists[l] < tmp) {
 99                     tmp = exists[l];
100                     min = l;
101                 }
102             }
103         }
104         //用序号标记走过的位置
105         i = nextX[min];
106         j = nextY[min];
107         pos[i][j] = m;
108     }
109     return 1;
110 }