第十屆藍橋杯大賽軟體類省賽 迷宮問題題解
阿新 • • 發佈:2020-09-16
問題描述
從左上角到右下角的路徑。上下左右分別用UDLR表示,求用UDLR表示的路徑。 輸入資料: 30 50 01010101001011001001010110010110100100001000101010 00001000100000101010010000100000001001100110100101 01111011010010001000001101001011100011000000010000 01000000001010100011010000101000001010101011001011 00011111000000101000010010100010100000101100000000 11001000110101000010101100011010011010101011110111 00011011010101001001001010000001000101001110000000 10100000101000100110101010111110011000010000111010 00111000001010100001100010000001000101001100001001 11000110100001110010001001010101010101010001101000 00010000100100000101001010101110100010101010000101 11100100101001001000010000010101010100100100010100 00000010000000101011001111010001100000101010100011 10101010011100001000011000010110011110110100001000 10101010100001101010100101000010100000111011101001 10000000101100010000101100101101001011100000000100 10101001000000010100100001000100000100011110101001 00101001010101101001010100011010101101110000110101 11001010000100001100000010100101000001000111000010 00001000110000110101101000000100101001001000011101 10100101000101000000001110110010110101101010100001 00101000010000110101010000100010001001000100010101 10100001000110010001000010101001010101011111010010 00000100101000000110010100101001000001000000000010 11010000001001110111001001000011101001011011101000 00000110100010001000100000001000011101000000110011 10101000101000100010001111100010101001010000001000 10000010100101001010110000000100101010001011101000 00111100001000010000000110111000000001000000001011 10000001100111010111010001000110111010101101111000
題解思路:BFS
1.從終點向原點使用BFS搜尋
2.定義一個dist[][]陣列,表示某一點從終點向原點的距離
3.從dist[][]陣列的原點往終點求路徑(不能再向終點求原點的距離)
C/C++實現
#define MAX_VALUE 100 // 迷宮問題 string maze[MAX_VALUE]; int dist[MAX_VALUE][MAX_VALUE]; int dire[4][2] = { {1,0}, //down {0,-1}, // left {0,1}, // right {-1,0} // up }; char dir[4] = {'D','L','R','U'}; queue<pair<int,int>>Q; bool check(int x,int y, int m,int n){ return (x >= 0 && y >= 0 && x < m && y < n && dist[x][y] == -1 && maze[x][y] == '0'); } void maze_bfs(int m,int n , int x_start,int y_start){ memset(dist,-1, sizeof(dist)); Q.push(make_pair(x_start,y_start)); dist[x_start][y_start] = 0; while(Q.size()){ pair<int,int> pos = Q.front(); Q.pop(); for(int i = 0;i < 4;i++){ int new_x = pos.first + dire[i][0]; int new_y = pos.second + dire[i][1]; if (check(new_x, new_y, m,n)){ dist[new_x][new_y] = dist[pos.first][pos.second]+1; Q.push(make_pair(new_x,new_y)); } } } } void p_maze(){ int m,n; cin >> m >> n; //memset(maze,-1,sizeof(maze)); for(int i = 0;i < m;i++) cin >> maze[i]; maze_bfs(m,n,29,49); } int main() { p_maze(); cout << endl; for(int i = 0;i < 30;i++) { for (int j = 0; j < 50; j++) if (dist[i][j] != -1) printf("%3d ",dist[i][j]); else printf(" "); cout << endl; } int x = 0,y = 0; int d = dist[x][y]; string ret; while(d){ for (int i = 0; i < 4; ++i) { int temp_x = x+dire[i][0]; int temp_y = y+dire[i][1]; if((d-1) == dist[temp_x][temp_y]){ ret += dir[i]; x = temp_x; y = temp_y; d--; break; } } } cout << ret; return 0; }