UVa 532 - Dungeon Master

　　题目大意：给一个三维迷宫，给出入口和出口，找出最短路径。

　　无权图上的单源最短路问题，使用BFS解决。

 #include <cstdio>
 #include <queue>
 #include <cstring>
 using namespace std;
 #define MAXN 35

 const int dir[][] = {{-, }, {, -}, {, }, {, }, {-, }};
 int G[MAXN][MAXN][MAXN], dist[MAXN*MAXN*MAXN];
 int L, R, C;
 int s, e;

 int bfs()
 {
     queue<int> q;
     memset(dist, -, sizeof(dist));
     q.push(s);
     dist[s] = ;
     while (!q.empty())
     {
         int u = q.front();
         q.pop();
         int l = u / (R*C);
         int r = (u%(R*C)) / C;
         int c = (u%(R*C)) % C;
         for (int i = ; i < ; i++)
         {
             int rr = r + dir[i][];
             int cc = c + dir[i][];
             int v = l*(R*C) + rr*C + cc;
             if (rr >=  && rr < R && cc >=  && cc < C && G[l][rr][cc] && dist[v] == -)
             {
                 q.push(v);
                 dist[v] = dist[u] + ;
                 if (v == e)  return dist[e];
             }
         }
         for (int i = ; i < ; i++)
         {
             int ll = l + dir[][i];
             int v = ll*(R*C) + r*C + c;
             if (ll >=  && ll < L && G[ll][r][c] && dist[v] == -)
             {
                 q.push(v);
                 dist[v] = dist[u] + ;
                 if (v == e)  return dist[e];
             }
         }
     }
     return dist[e];
 }

 int main()
 {
 #ifdef LOCAL
     freopen("in", "r", stdin);
 #endif
     char str[];
     while (scanf("%d%d%d", &L, &R, &C) && (L || R || C))
     {
         getchar();
         for (int k = ; k < L; k++)
         {
             for (int i = ; i < R; i++)
             {
                 gets(str);
                 for (int j = ; j < C; j++)
                 {
                     if (str[j] == '.')  G[k][i][j] = ;
                     else if (str[j] == '#')  G[k][i][j] = ;
                     else if (str[j] == 'S')
                     {
                         s = k*(R*C) + i*C + j;
                         G[k][i][j] = ;
                     }
                     else if (str[j] == 'E')
                     {
                         e = k*(R*C) + i*C + j;
                         G[k][i][j] = ;
                     }
                 }
             }
             gets(str);
         }
         int ans = bfs();
         if (ans > )  printf("Escaped in %d minute(s).\n", ans);
         else if (ans == -)  printf("Trapped!\n");
     }
     return ;
 }