数据结构-迷宫

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// 迷宫问题.cpp : 定义控制台应用程序的入口点。 #include "stdafx.h" #define M 4 #define N 4 #define MaxSize 100 int  mg[M + 2][N + 2] = {     {1,1,1,1,1,1},     {1,0,0,0,0,1},     {1,0,1,0,0,1},     {1,1,0,1,0,1},     {1,0,0,0,0,1},     {1,1,1,1,1,1} };   struct {    int i;int  j; int di; }Stack[MaxSize],Path[MaxSize]; // 定义栈和最短路径的数组   // 初始化 int top = -1; int count = 1; int minlen = MaxSize;   void mgpath() {     int i, j, di, find, k;     top++;     Stack[top].i = 1;     Stack[top].j = 1;     Stack[top].di = -1;     mg[1][1] = -1;       //所有方向走完，回到初始点，退出栈，站内为空     while (top>-1)     {         i = Stack[top].i;         j = Stack[top].j;         di = Stack[top].di;           // 走出迷宫，打印当前路线信息，比较路径长度，找出最短路径。         if (i == M && j == N)         {             printf_s("%4d:", count++);             for (k = 0; k <= top; k++)             {                 printf_s("(%d,%d) ", Stack[k].i, Stack[k].j);                 if ((k + 1) % 5 == 0)                 {                     printf_s("\n\t");                 }             }             printf_s("\n");             if (top + 1 < minlen)             {                 for (k = 0; k <= top; k++)                 {                     Path[k] = Stack[k];                 }                 minlen = top + 1;             }                 // 开始回溯路径，寻找其他路径             mg[Stack[top].i][Stack[top].j] = 0;             top--;             i = Stack[top].i;             j = Stack[top].j;             di = Stack[top].di;         }           find = 0;         while (di<4&&find==0)         {             di++;             // 迷宫方向选择（上、右、下、左）             switch (di)             {             case 0:  // 向上                 i = Stack[top].i - 1;                 j = Stack[top].j;                 break;             case 1: // 向右                 i = Stack[top].i;                 j = Stack[top].j + 1;                 break;             case 2: // 向下                 i = Stack[top].i + 1;                 j = Stack[top].j;                 break;             case 3: // 向左                 i = Stack[top].i;                 j = Stack[top].j - 1;                 break;             }             if (mg[i][j] == 0)             {                 find = 1;               }         }           if (find == 1)              // 找到下一个可走节点         {             Stack[top].di = di;     // 更改原栈顶元素的方向di值，小于di方向已走，大于di方向未走。             top++;                  // 可走节点入栈，             Stack[top].i = i;             Stack[top].j = j;             Stack[top].di = -1;     // 可走节点方向从-1开始到3进行判断下一个可走节点             mg[i][j] = -1;          //避免重复走到节点         }         else                        // 没有路径可走，则退栈（所有信息下次入栈时重新初始化）         {             mg[Stack[top].i][Stack[top].j] = 0;         //最后初始节点所有方向走完，无可走节点，退栈，栈空             top--;         }     }       // 打印迷宫最短路径。     if (minlen >= MaxSize)     {         printf_s("迷宫没有出口：\n");     }     else     {         printf_s("最短路径如下：\n");         printf_s("长度:%d \n", minlen);         for (k = 0; k < minlen; k++)         {             printf_s("(%d,%d)", Path[k].i, Path[k].j);             if ((k + 1) % 5 == 0)             {                 printf_s("\n\t");             }         }     } }   int main() {          int i=0;     mgpath();     scanf_s("%d", i);     return 0; }