ACM: Figure BFS, Maze

Title:

Network Maze n rows M unit columns, each cell regardless of space (using 1 representation), whether it is a barrier (using 0 in order to represent). Your task is to find an action sequence that is the shortest from start to finish, where the UDLR increases, the next, left, and right moves to the next cell.

No matter what time you are in the barrier. Can not go beyond the maze.

The starting and ending points are guaranteed to be open spaces.

Analysis: The BFS of the figure.

#include <iostream> #include <string> #include <queue>using namespace std;const int maxn = 500;int maze[ MAXN][MAXN], VIS[MAXN][MAXN], DIST[MAXN][MAXN], FA[MAXN][MAXN], Last_dir[maxn][maxn];int N, M, XS, Ys, XT, yt;int dx[] = { -1, 1, 0, 0};int dy[] = {0, 0,-1, 1};char name[] = "UDLR"; void Print_path (int x, int y) {//recursively print path int FX = Fa[x]    [y]/m;    int fy = fa[x][y]% m;    if (FX! = x | | fy! = y) {print_path (FX, FY);p Utchar (Name[last_dir[x][y]); }}int dir[maxn*maxn];void print_path2 (int x, int y) {//To print the path in an iterative way int c = 0;for (;;) {int FX = Fa[x][y]/m;int fy = fa[x][y]% m;if (FX = = x && fy = = y) break;dir[c++] = Last_dir[x][y];x = FX; y = fy ;} while (c--) Putchar (Name[dir[c]);}   queue<int> q;void BFS (int x, int y) {int u = x*m+y;dist[x][y] = 0;     Initialize yourself to your own distance is 0fa[x][y] = u; The Father node of the starting point is himself.   Easy to follow the printing operation vis[x][y] = 1; Q.push (U), while (!q.empty ()) {u = Q.front (); Q.pop (); x = U/m;y = u%m;for (int d = 0; d < 4; ++d) {int NX = x + dx[d];int n y = Y + dy[d];if (NX >= 0 && NX < n && NY >= 0 && NY < m && Maze[nx][ny] &&amp    ;!vis[nx][ny]) {int v = NX * m + ny;q.push (v); Vis[nx][ny] = 1;dist[nx][ny] = Dist[x][y] + 1;                   Number of steps to walk +1fa[nx][ny] = v;             Record father knot Point last_dir[nx][ny] = D; Record the direction of this node to Father's node now}}}}int main () {cin >> n >> M >> xs >> ys >> xt >> yt;for (int i = 0 ; I < n; ++i) {for (int j = 0; j < m; ++j) {cin >> maze[i][j];}} memset (Vis, 0, sizeof (VIS)), BFS (XS, YS);p Rint_path (XT, YT), cout << endl;print_path2 (XT, YT); cout << Endl; return 0;}

ACM: Figure BFS, Maze