int initgame(HWND hWnd); int markpath(int x, int y, int dir, int color); int drawWall(int x, int y, int dir); int drawMaze(); int readMazedata(int *num, int id); char *GetFldStr(char *sFldStr, char *sStr, char sFld); void initMalloc(); void addMallocNode(char *); ...
(); // 初始化游戏图片 void InitGame(); // 初始化游戏数据 void GetMazeSize(); // 提示用户输入迷宫大小 void MakeMaze(int width, int height); // 生成迷宫:初始化(注:宽高必须是奇数) void TravelMaze(int x, int y); // 生成迷宫:遍历 (x, y) 四周 MAPITEM GetMazeItem(int x, int ...
int init(int** Maze); //初始化迷宫 void print(int** Maze);//画迷宫 void CreateMaze(int **maze, int x, int y); //创建迷宫 void move(int** Maze, char t, int *x, int *y);//移动角色 int main() { SetConsoleTitle("ytr的迷宫test"); menu(); return 0; } void HideCursor() ...
// 初始化voidWelcome();// 绘制游戏界面voidReadyGo();// 准备开始游戏voidInitGame();// 初始化游戏数据// 矿井生成voidMakeMaze(intwidth,intheight);// 初始化(注:宽高必须是奇数)voidTravelMaze(intx,inty,BYTE**aryMap);// 遍历 (x, y) 四周voidDrawWall(intx,inty,boolleft,booltop,boolright...
CF1293C - NEKO's Maze Game 分块 一定是两个障碍物组成一对来破坏连通性,每个障碍物可能属于最多3对,然后维护障碍物对数就行。但是懒得讨论,暴力分块过了。 涉及到修改的块暴力重构这个块的连通性。只要左端两个位置和右端两个位置中任意两个可互达就具有连通性。
CMD_MARKYELLOW=128,CMD_CLEARMARK=256};/// 函数声明//voidWelcome();// 绘制游戏界面voidInitImage();// 初始化游戏图片voidInitGame();// 初始化游戏数据voidGetMazeSize();// 提示用户输入迷宫大小voidMakeMaze(intwidth,intheight);// 生成迷宫:初始化(注:宽高必须是奇数)voidTravelMaze(intx,inty);...
AC代码: #include <cstdio> #include <vector> #include <queue> #include <cstring> #include <cmath> #include #include <set> #include <string> #include <iostream> #include <algorithm> #include <iomanip> #include <stack> #include <queue> using ...
Referee and state: The rule-enforcer and game data. PlayerMechanism: Represents the basic functionalities a player supports (setup, take-turn, win) Server and Client: Allows remote interactions between remote players and the game server Maze/Client contains implementation of the client Maze/Common...
void InitGame(); // 初始化游戏数据 // 矿井生成 void MakeMaze(int width, int height); // 初始化(注:宽高必须是奇数) void TravelMaze(int x, int y, BYTE** aryMap); // 遍历 (x, y) 四周 void DrawWall(int x, int y, bool left, bool top, bool right, bool bottom); ...
C. NEKO's Maze Game 题目链接: C. NEKO’s Maze Game time limit per test:1.5 seconds memory limit per test:256 megabytes inputstandard input outputstandard output NEKO#ΦωΦ has just got a new maze game on her PC! The game’s main puzzle is a maze, in the forms of a 2×n ...