纯c语言迷宫源码.zip
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纯c语言迷宫源码.zip #include <stdio.h>#include <stdlib.h>#include <time.h>int maze[100][100];int m=0,n=0;/**对迷宫进行初始化，用随机数产生迷宫*/void InitMaze(){ int i,j,temp; srand((unsigned)time(NULL)); for(i=1;i<=m;i++) for(j=1;j<=n;j++) { temp=rand()%100; if(temp>30) { maze[i-1][j-1]=0; }else { maze[i-1][j-1]=1; } } maze[0][0]=0; maze[m-1][n-1]=9;}/**定义栈和栈的节点*/typedef struct Node{int x;int y;struct Node *next;}Node,*Stack;/**初始化Stack*/void InitStack(Node *Stack){Stack=(Node *)malloc(sizeof(Node));if(Stack==NULL){ printf("分配空间失败\n"); exit(0);}else{ Stack->next=NULL;}}/**压栈*/void push(Node *Stack,int x,int y){Node *temp;temp=(Node *)malloc(sizeof(Node));if (!temp){ printf("分配内存空间错误"); return;}else{ temp->x=x; temp->y=y; temp->next=Stack->next; Stack->next=temp;}}/**出栈*/void pop(Node *Stack,int *x,int *y){Node *temp;temp=Stack->next;if(!temp){ return;}else{ *x=temp->x; *y=temp->y;}Stack->next=temp->next;free(temp);}/**判断栈是否为空*/int isEmpty(Node *Stasck){ return ((Stasck->next)==NULL);}/**判断从该点时候可以向其他方向移动，并返回移动的方向*/int pass(int i,int j){ if(j<n-1&&(maze[i][j+1]==0||maze[i][j+1]==9)) { return 2; } if(i<m-1&&(maze[i+1][j]==0||maze[i+1][j]==9)) { return 3; } if(j>=1&&(maze[i][j-1]==0||maze[i][j-1]==9)) { return 4; } if(i>=1&&(maze[i-1][j]==0||maze[i-1][j]==9)) { return 5; } return -1;}/**对迷宫进行打印*/void drawMaze(){ int i=0,j=0; for(i=0;i<m;i++){ for(j=0;j<n;j++) { if(maze[i][j]==0||maze[i][j]==9||maze[i][j]==-1) { printf("a"); } else if(maze[i][j]==1) { printf("b"); } else if(maze[i][j]==2) { printf("D"); }else if(maze[i][j]==3) { printf("X"); }else if(maze[i][j]==4) { printf("A"); }else if(maze[i][j]==5) { printf("W"); } } printf("\n"); }}/**对迷宫进行路径搜索*数组的数字有以下含义*0.该点没有被探索过，且可行*1.该点不可行*2.该点是可行的，且进行了向东的探索*3.该点是可行的，且进行了向南的探索*4.该点是可行的，且进行了向西的探索*5.该点是可行的，且进行了向北的探索*6.该点是入口*9.该点是出口*-1.该点已经遍历完毕四个方向，不能找到有效的路径，则置为－1*/void ShowPath(){int curx=0,cury=0;int count=0;int flag=0;Node *Stacks=NULL;InitStack(Stacks);do{ if(maze[curx][cury]==9) { flag=1; } switch(pass(curx,cury)){ case 2: maze[curx][cury]=2; push(Stacks,curx,cury); cury++; break; case 3: maze[curx][cury]=3; push(Stacks,curx,cury); curx++; break; case 4: maze[curx][cury]=4; push(Stacks,curx,cury); cury--; break; case 5: maze[curx][cury]=5; push(Stacks,curx,cury); curx--; break; case -1: maze[curx][cury]=-1; if(!isEmpty(Stacks)) pop(Stacks,&curx,&cury); break; } count++;}while(!isEmpty(Stacks)&&flag==0);if(flag==1){ printf("The path is :\n"); printf("\n"); drawMaze();}else{ printf("\nSorry,you fail\n");}}/**主函数*/int main(){ loop: printf("plase input the number of row m(m>0,m<100):"); scanf("%d",&m); printf("plase input the number of line n(n>0,n<100):"); scanf("%d",&n); if(m<0||m>100||n<0||n>100){ printf("The number is error,process will exit !\n"); exit(-1); } printf("The character is 'a',it is area.\n"); printf("The character is 'b',it is wall.\n"); printf("\n"); InitMaze(); printf("The oid Maze:\n"); printf("\n"); drawMaze(); printf("\n show the path ?(y/n)"); fflush(stdin); if(toupper(getch())=='Y') { printf("\n"); ShowPath(); printf("\n go on play ?(y/n)"); fflush(stdin); if(toupper(getch())=='Y') { goto loop; } else { exit(1); } } else { exit(1); } getch(); return 0;}