机器博弈中的数据结构与基本方法（二）-----递归及回溯法实例
1、跳马问题：从左上角开始，按照象棋中马的行走规则（但是不考虑马脚，只需要按日字走），要求每个点走一次且仅一次，并且所有的点走走到，求解所有可行走法。
思路及代码：
1 #include <cstdio> 2 int a[9][9]; //包含边界拓展 3 int S = 0; //总的方案数 4 void PrintAnswer(){ //打印行走策略 5 printf("Answer\t%d\n", S); 6 for (int i = 2; i < 7; i++){ 7 for (int j = 2; j < 7; j++){ 8 printf("%-4d\t", a[i][j]); 9 } 10 printf("\n"); 11 } 12 } 13 void Try(int i, int ml, int nl); 14 void Move(int i, int m, int n){ 15 if (0 == a[m][n]){ 16 a[m][n] = i; 17 if (i < 25){ 18 Try(i+1, m, n); 19 } else { 20 S++; 21 PrintAnswer(); 22 } 23 a[m][n] = 0; 24 } 25 } 26 void Try(int i, int ml, int nl){ //i：到达结点编号，ml/nl出发点的位置 27 int n, m; 28 m = ml - 2; 29 n = nl + 1; //走法 1 30 Move(i, m, n); 31 32 m = ml - 2; 33 n = nl - 1; //走法 2 34 Move(i, m, n); 35 36 m = ml + 2; 37 n = nl + 1; //走法 3 38 Move(i, m, n); 39 40 m = ml + 2; 41 n = nl - 1; //走法 4 42 Move(i, m, n); 43 44 m = ml - 1; 45 n = nl + 2; //走法 5 46 Move(i, m, n); 47 48 m = ml - 1; 49 n = nl - 2; //走法 6 50 Move(i, m, n); 51 52 m = ml + 1; 53 n = nl + 2; //走法 7 54 Move(i, m, n); 55 56 m = ml + 1; 57 n = nl - 2; //走法 8 58 Move(i, m, n); 59 } 60 int main(int argc, char const *argv[]) 61 { 62 freopen("ans.txt", "w", stdout); 63 for (int i = 0; i < 9; i++){ 64 for (int j = 0; j < 9; j++){ 65 if (i < 2 || i > 6 || j < 2 || j > 6){ 66 a[i][j] = -1; 67 } else { 68 a[i][j] = 0; 69 } 70 } 71 } 72 a[2][2] = 1; 73 Try(2, 2, 2); 74 return 0; 75 }
部分答案
1 Answer 301 2 1 6 17 12 25 3 16 11 2 7 18 4 3 20 5 24 13 5 10 15 22 19 8 6 21 4 9 14 23 7 Answer 302 8 1 16 7 12 25 9 8 13 2 17 6 10 3 20 15 24 11 11 14 9 22 5 18 12 21 4 19 10 23 13 Answer 303 14 1 18 7 12 25 15 8 13 2 19 6 16 3 20 17 24 11 17 14 9 22 5 16 18 21 4 15 10 23 19 Answer 304 20 1 8 19 14 25 21 18 13 2 9 6 22 3 20 7 24 15 23 12 17 22 5 10 24 21 4 11 16 23 25 共 304 种策略
图示
2、迷宫问题：如下图，求出所有可以到达出口的路径（左上角绿色是入口，右下角是出口，深红色部分是墙，红色是为了编程方便（边界判断）人为加的）。
代码及思路
1 #include <cstdio> 2 int a[8][12]; 3 int s = 0; 4 void PrintPath(){ 5 printf("Path%d\n", s); 6 for (int i = 1; i < 7; i++){ 7 for (int j = 1; j < 11; j++) 8 { 9 printf("%4d", a[i][j]); 10 } 11 printf("\n"); 12 } 13 } 14 void Try(int i, int ml, int nl); 15 void Move(int i, int m, int n){ 16 if (0 == a[m][n]){ 17 a[m][n] = i; 18 if (6 == m && 10 == n){ 19 s++; 20 PrintPath(); 21 } else { 22 Try(i+1, m, n); 23 } 24 a[m][n] = 0; 25 } 26 } 27 28 void Try(int i, int ml, int nl){ 29 int m, n; 30 m = ml; 31 n = nl - 1; //注意只能上下左右与移动 32 Move(i, m, n); 33 34 m = ml; 35 n = nl + 1; 36 Move(i, m, n); 37 38 m = ml - 1; 39 n = nl; 40 Move(i, m, n); 41 42 m = ml + 1; 43 n = nl; 44 Move(i, m, n); 45 } 46 int main(int argc, char const *argv[]) 47 { 48 freopen("ans.txt", "w", stdout); 49 for (int i = 0; i < 8; i++){ 50 for (int j = 0; j < 12; j++){ 51 a[i][j] = -1; 52 } 53 } 54 a[1][1] = 0; 55 for (int i = 1; i <= 8; i++){ 56 if (2!=i && 5!=i){ 57 a[2][i] = 0; 58 } 59 } 60 for (int i = 1; i <= 9; i++){ 61 if (4!=i && 5!=i && 7!=i){ 62 a[3][i] = 0; 63 } 64 } 65 for (int i = 1; i <= 9; i++){ 66 if (2!=i && 3!=i && 7!=i && 8!=i){ 67 a[4][i] = 0; 68 } 69 } 70 for (int i = 1; i <= 9; i++){ 71 if (5!=i){ 72 a[5][i] = 0; 73 } 74 } 75 a[6][9] = a[6][10] = 0; 76 a[1][1] = 1; 77 Try(2, 1, 1); 78 printf("Sum = %d\n", s); 79 return 0; 80 }
输出结果
Path1 1 -1 -1 -1 -1 -1 -1 -1 -1 -1 2 -1 0 0 -1 13 14 15 -1 -1 3 0 0 -1 -1 12 -1 16 17 -1 4 -1 -1 9 10 11 -1 -1 18 -1 5 6 7 8 -1 0 0 0 19 -1 -1 -1 -1 -1 -1 -1 -1 -1 20 21 Path2 1 -1 -1 -1 -1 -1 -1 -1 -1 -1 2 -1 0 0 -1 0 0 0 -1 -1 3 0 0 -1 -1 0 -1 0 0 -1 4 -1 -1 9 10 11 -1 -1 0 -1 5 6 7 8 -1 12 13 14 15 -1 -1 -1 -1 -1 -1 -1 -1 -1 16 17 Sum = 2
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