本文实例讲述了C语言实现简单的走迷宫游戏的方法，代码完整，便于读者理解。

学数据结构时用“栈”写的一个走迷宫程序，实际上用到双向队列，方便在运行完毕后输出经过的点。

#include <cstdio> #include <deque> #include <windows.h> using namespace std; class node { public: int x,y; int lastOpt; }; deque<node> sta; int x,y; int endx,endy; int mapW,mapH; int steps; int xopt[5]= {0,0,1,0,-1}; int yopt[5]= {0,1,0,-1,0}; int map[100][100]= { }; void init() { x = 1; y = 1; endx = 1; endy = 9; mapH = 10; mapW = 10; for(int i=0; i<=mapH; i++) for(int j=0; j<=mapW; j++) { if(i==0 ||j==0 ||i==mapH||j==mapW) map[i][j]=-1; } steps=0; map[1][2]=-1; map[2][2]=-1; map[3][2]=-1; map[4][2]=-1; map[6][2]=-1; map[7][2]=-1; map[8][2]=-1; map[9][2]=-1; map[9][3]=-1; map[8][3]=-1; map[1][4]=-1; map[3][4]=-1; map[4][4]=-1; map[5][4]=-1; map[6][4]=-1; map[7][4]=-1; map[1][6]=-1; map[2][6]=-1; map[3][6]=-1; map[4][6]=-1; map[5][6]=-1; map[6][6]=-1; map[7][6]=-1; map[8][6]=-1; map[8][7]=-1; map[8][8]=-1; map[7][8]=-1; map[6][8]=-1; map[5][8]=-1; map[4][8]=-1; map[3][8]=-1; map[2][8]=-1; map[1][8]=-1; map[endx][endy]=5; } void dis() { system("cls"); int ori = map[x][y]; map[x][y]=1; for(int i=0; i<=mapH; ++i) { for(int j=0; j<=mapW; ++j) { if(map[i][j]==0) printf(" "); else if(map[i][j]==-1) printf(" #"); else if(map[i][j]==1) printf(" @"); else if(map[i][j]==2) printf(" ."); else if(map[i][j]==5) printf(" !"); } cout<<i<<endl; } for(int j=0; j<=mapW; ++j) cout<<j<<" "; printf("nn > steps:%d Exit:(%d,%d)n",steps,endx,endy); map[x][y] = ori; } int can(int n) { if(map[x+xopt[n]][y+yopt[n]] == 0 || map[x+xopt[n]][y+yopt[n]] == 5) return 1; return 0; } void visit(int n) { map[x][y]=2; x+=xopt[n]; y+=yopt[n]; node tem; tem.x = x; tem.y = y; tem.lastOpt = n; sta.push_back(tem); steps++; } int main() { init(); node tem; while( x != endx || y!=endy) { int cans = 0; for(int i=1; i<=4; i++) { if(can(i)) { cans = 1; visit(i); break; } } if(!cans) { if(!sta.empty()) { tem = sta.back(); map[tem.x][tem.y]=0; sta.pop_back(); } else { map[x][y]=2; x+=xopt[tem.lastOpt]; x+=yopt[tem.lastOpt]; dis(); break; } } dis(); Sleep(500); } if(x==endx && y == endy) cout<<"n > i am finished....n"; else cout<<"n > i am finished...but i can't find the right wayn"; return 0; }

效果图：

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