用回溯法求解N皇后和迷宫问题

关于回溯法不再赘述太多，大家移步各种百科，本文介绍两种应用的实现

其典型应用之一： N皇后问题

按皇后摆放规则顺序遍历棋盘依次安放皇后，直到某一个皇后找不到合适的位置时，倒退至上一步，调整上一个皇后位置，如还不满足，继续调整再上一个，依次类推。

上述过程很容易想到用栈实现，其一组特解的代码如下

void search_solution(){struct position tmp;int i = 0, j = 0;for (; i < N; ){for (; j < N; j++){if (detect(i,j)){map[i][j] = 1;push(i,j);break;}}if (j == N){tmp = pop();i = tmp.i;j = tmp.j;map[i][j] = 0;j++;}else{i++;j = 0;}}}

要找出所有解，在找到一组解后（遍历到最后一行满足条件）不退出，继续出栈，直到栈空（我的程序在设计时，直接指定了空栈时安全退出程序）：

void search_all_solution(){struct position tmp;int i = 0, j = 0;while (1){for (; i < N; ){for (; j < N; j++){if (detect(i,j)){map[i][j] = 1;push(i,j);break;}}if (j == N){tmp = pop();i = tmp.i;j = tmp.j;map[i][j] = 0;j++;}else{i++;j = 0;}}if (i == N){count++;print_solution();tmp = pop();i = tmp.i;j = tmp.j;map[i][j] = 0;j++;}}}

完整代码如下：

#include<stdio.h>#include<string.h>#define N 8#define MAX_STACK_SIZE 64struct position{int i, j;};int map[N][N];int empty_flag = 0;int top = -1;int count = 0;struct position stack[MAX_STACK_SIZE];void push(int i, int j){top++;stack[top].i = i;stack[top].j = j;empty_flag = 0;}struct position pop(){if (top == -1){printf("%d solutions\n", count);exit(0);printf("stack is empty!\n");empty_flag = 1;}return stack[top--];}void print_solution(){int i, j;for (i = 0; i < N; i++){for (j = 0; j < N; j++)printf("%2d", map[i][j]);printf("\n");}printf("\n");}int detect(int row, int col){int i = row, j;for (j = 0; j < N; j++)if (map[i][j])return 0;j = col;for (i = 0; i < N; i++)if (map[i][j])return 0;i = row;while (++i < N && ++j < N){if (map[i][j])return 0;}i = row;j = col;while (--i >= 0 && --j >= 0){if (map[i][j])return 0;}i = row;j = col;while (++i < N && --j >= 0){if (map[i][j])return 0;}i = row;j = col;while (--i >= 0 && ++j < N){if (map[i][j])return 0;}return 1;}void search_solution(){struct position tmp;int i = 0, j = 0;for (; i < N; ){for (; j < N; j++){if (detect(i,j)){map[i][j] = 1;push(i,j);break;}}if (j == N){tmp = pop();i = tmp.i;j = tmp.j;map[i][j] = 0;j++;}else{i++;j = 0;}}}void search_all_solution(){struct position tmp;int i = 0, j = 0;while (1){for (; i < N; ){for (; j < N; j++){if (detect(i,j)){map[i][j] = 1;push(i,j);break;}}if (j == N){tmp = pop();i = tmp.i;j = tmp.j;map[i][j] = 0;j++;}else{i++;j = 0;}}if (i == N){count++;print_solution();tmp = pop();i = tmp.i;j = tmp.j;map[i][j] = 0;j++;}}}int main(){memset(map, 0, sizeof(map));//search_solution();//print_solution();search_all_solution();printf("%d solutions\n", count);return 0;}

应用之二： 走迷宫

遇到分岔路口就把分岔的坐标入栈，沿着其中一条走下去，如果无解返回分岔坐标，如此类推：

解释下程序中的矩阵，0表示路，1表示墙，走过的路径用8表示。每走一步打印一次地图可观察到程序的执行情况

#include<stdio.h>#include<string.h>#define ROW 7#define COL 7#define MAX_STACK_SIZE 32#define UP0#define RIGHT   1#define DOWN    2#define LEFT    3int maze[ROW][COL] = {  0,1,0,0,0,1,0,  0,1,1,1,0,0,0,  0,0,0,1,0,1,0,  0,1,1,1,0,1,1,  0,0,0,1,0,0,0,  0,1,0,1,0,1,0,  0,0,0,0,0,1,0,};typedef struct Point_tag{int row, col;}Point;Point stack[MAX_STACK_SIZE];int top = -1;int direction[4] = {0, 0, 0, 0}; //从第一个元素开始，依次标识上（0），右（1），下(2)，左(3), 表示方向貌似有更好的方法void push(Point po){if (top == MAX_STACK_SIZE - 1)printf("stack is full!\n");elsestack[++top] = po; }Point pop(){if (top == -1)printf("stack is empty!\n");elsereturn stack[top--];}void print_maze(){int i, j;for (i = 0; i < ROW; i++){for (j = 0; j < COL; j++)printf("%2d", maze[i][j]);printf("\n");}printf("\n");}void count_direction(Point po, int current_dir){if (po.row + 1 < ROW && maze[po.row+1][po.col] != 1 && maze[po.row+1][po.col] != 8 /*&& current_dir != UP*/)direction[DOWN] = 1;if (po.col + 1 < COL && maze[po.row][po.col+1] != 1 && maze[po.row][po.col+1] != 8 )direction[RIGHT] = 1;if (po.row - 1 >= 0 && maze[po.row-1][po.col] != 1 && maze[po.row-1][po.col] != 8)direction[UP] = 1;if (po.col - 1 >= 0 && maze[po.row][po.col-1] != 1 &&  maze[po.row][po.col-1] != 8)direction[LEFT] = 1;}int num_of_dir(){int i = 0, count = 0;for (; i < 4; i++)if (direction[i] == 1)count++;return count;}int local_dir(){int i = 0;for (; i < 4; i++)if (direction[i] == 1)return i;}void search(Point head , int cur_dir){int numofdir, next_dir = 0, dir = cur_dir;while (1){maze[head.row][head.col] = 8;count_direction(head, dir);numofdir = num_of_dir();if (numofdir == 0)head = pop();else {if (numofdir > 1)push(head);next_dir = local_dir();switch (next_dir){case UP: { head.row--; break; }case RIGHT: { head.col++; /*dir = RIGHT;*/ break; }case DOWN: { head.row++; break; }case LEFT: { head.col--; break; }}}memset(direction, 0, sizeof(int) * 4);print_maze();if (head.row == ROW - 1 && head.col == COL - 1){maze[head.row][head.col] = 8;print_maze();printf("success to escape!\n");break;}}}int main(){Point Head = {0, 0};int dir = 2;search(Head, dir);return 0;}

程序执行情况部分截图：