走迷宫问题(待续)
前言:
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我的地图文件(MazeMap.txt)如下:
size:(a,a) 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1
下面的pos类用来保存某个位置的坐标
GetMaze函数用来判断地图格式的合法性,若合法则读取地图内容,并将内容存入参数arr所指向的内存中。
struct pos { pos(int row = 0, int col = 0) :_row(row) ,_col(col) {} int _row; int _col; }; void GetMaze(int *&arr, int &sz, int &row, int &col) { FILE *fout = fopen("MazeMap.txt", "r"); assert(fout); char *title = new char[40]; title = fgets(title, 7, fout); if (strcmp(title, "size:(")) { cout << "地图文件错误!" << endl; throw 1; } row = fgetc(fout) - 87; title = fgets(title, 2, fout); if (strcmp(title, ",")) { cout << "地图文件错误!" << endl; throw 1; } col = fgetc(fout) - 87; arr = new int[row * col]; sz = row * col; title = fgets(title, 2, fout); for (int i = 0; i < sz; i) { char ch = fgetc(fout); if (ch != ' ' && ch != '\n' && ch != '\0') { *(arr + i) = ch - '0'; i++; } } }
一、找到出口
bool MazePath(int *arr, int n, const pos &entry, stack&path) //假设下边沿为迷宫的出口 { pos cur = entry; path.push(cur); while (!path.empty()) { *(arr + n * (cur._row)+cur._col) = 2; if (cur._row == n - 1) { return true; } //向下 if ((cur._row + 1 < 10) && (*(arr + n * (cur._row + 1) + cur._col) == 0)) { *(arr + n * (cur._row + 1) + cur._col) = 2; ++cur._row; path.push(cur); continue; } //向上 if ((cur._row - 1 >= 0) && (*(arr + n * (cur._row - 1) + cur._col) == 0)) { *(arr + n * (cur._row - 1) + cur._col) = 2; --cur._row; path.push(cur); continue; } //向左 if ((cur._col - 1 >= 0) && (*(arr + n * cur._row + cur._col - 1) == 0)) { *(arr + n * cur._row + cur._col - 1) = 2; --cur._col; path.push(cur); continue; } //向右 if ((cur._col + 1 < 10) && (*(arr + n * cur._row + cur._col + 1) == 0)) { *(arr + n * cur._row + cur._col + 1) = 2; ++cur._col; path.push(cur); continue; } //走不通 cur._col = path.top()._col; cur._row = path.top()._row; path.pop(); } }
二、找到所有出口并得出最短路线(最优解)
templatevoid ClearPath(stack &stack) { while (!stack.empty()) { stack.pop(); } } static void SaveBestPath(stack &path, vector< stack > path_vec) { stack best_path; int sz = (path_vec.back()).size(); best_path = path_vec.back(); while (!path_vec.empty()) { if (sz > (path_vec.front()).size()) { best_path = path_vec.front(); } path_vec.pop_back(); } path = best_path; } static struct Exit { Exit(int row, int col) :_row(row) ,_col(col) {} int _row; int _col; }; //堵住已知的出口 (为了不销毁数据,不使用引用) static void BlockKnownExit(int *arr, vector< Exit> exit_vec, int n) { Exit ext1 = exit_vec.back(); while (!exit_vec.empty()) { ext1 = exit_vec.back(); *(arr + n * ext1._row + ext1._col) = 3; exit_vec.pop_back(); } } //假设下边沿为迷宫的出口 bool MazePathMin(int *arr, int n, const pos &entry, stack &path) { vector< stack > path_vec; //用于存放所有的路径 vector< Exit > exit_vec; //用于存储所有出口 pos cur = entry; path.push(cur); while (!path.empty() ) { *(arr + n * (cur._row) + cur._col) = 2; if (cur._row == n - 1) { //找到一个出口 *(arr + n * (cur._row) + cur._col) = 3; Exit ext_tmp(cur._row, cur._col); path_vec.push_back(path); exit_vec.push_back(ext_tmp); //清空路径,寻找除该出口外的其他出口 ClearPath(path); GetMaze(arr, n); BlockKnownExit(arr, exit_vec, n); cur = entry; path.push(cur); } //向下 if ((cur._row + 1 < 10) && (*(arr + n * (cur._row + 1) + cur._col) == 0)) { *(arr + n * (cur._row + 1) + cur._col) = 2; ++cur._row; path.push(cur); continue; } //向上 if ((cur._row - 1 >= 0) && (*(arr + n * (cur._row - 1) + cur._col) == 0)) { *(arr + n * (cur._row - 1) + cur._col) = 2; --cur._row; path.push(cur); continue; } //向左 if ((cur._col - 1 >= 0) && (*(arr + n * cur._row + cur._col - 1) == 0)) { *(arr + n * cur._row + cur._col - 1) = 2; --cur._col; path.push(cur); continue; } //向右 if ((cur._col + 1 < 10) && (*(arr + n * cur._row + cur._col + 1) == 0)) { *(arr + n * cur._row + cur._col + 1) = 2; ++cur._col; path.push(cur); continue; } //走不通的时候: cur._col = path.top()._col; cur._row = path.top()._row; path.pop(); } //path为空 SaveBestPath(path, path_vec); //把最佳的路径保存进path中(仍然倒序) return (!path.empty()); }
三、优化后的算法
四、用递归实现
//待续
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