POJ-2251.DungeonMaster(三维BFS)

做题时需要注意，爬楼有向上和向下爬之分...

　　本题大意：输入 l, r, c, 分别代表地牢的楼层数和每层地牢的长和宽，地牢由rock and point and source and key组成，你初始在s位置，你只能向身边的四个方向和上下方向移动，问你是否能走出地牢，能的话求出最短路径。

　　本题思路：BFS爆搜就OK，我不喜欢存运动方向，所以用循环判断了，按道理循环较慢...所以根据读者喜好选择...

　　参考代码：

 #include <cstdio>
 #include <iostream>
 #include <queue>
 #include <cmath>
 #include <cstring>
 using namespace std;

 struct node {
     int x, y, z, step;
 } now, Next, source, key;
 const int maxn =  + ;
 int l, r, c, ans;
 char dungeon[maxn][maxn][maxn];
 bool vis[maxn][maxn][maxn];

 int bfs() {
     memset(vis, false, sizeof vis);
     source.step = ;
     vis[source.x][source.y][source.z] = true;
     queue <node> Q;
     Q.push(source);
     while(!Q.empty()) {
         now = Q.front();
         Q.pop();
         if(now.x == key.x && now.y == key.y && now.z == key.z)   return now.step;
         Next.step = now.step + ;
         for(int i = -; i <= ; i ++)
             if(i != ) {
                 Next.x = now.x + i, Next.y = now.y, Next.z = now.z;
                 if(Next.x >=  && Next.y >=  && Next.z >=  && Next.x < l && Next.y < r && Next.z < c && !vis[Next.x][Next.y][Next.z] && dungeon[Next.x][Next.y][Next.z] != '#' ) {
                     Next.step = now.step + ;
                     vis[Next.x][Next.y][Next.z] = true;
                     Q.push(Next);
                 }
             }
         for(int dy = -; dy <= ; dy ++) {
             for(int dz = -; dz <= ; dz ++) {
                 if((int)(abs(dy - dz)) == ) {
                     Next.x = now.x + , Next.y = now.y + dy, Next.z = now.z + dz;
                     if(Next.x >=  && Next.y >=  && Next.z >=  && Next.x < l && Next.y < r && Next.z < c && !vis[Next.x][Next.y][Next.z] && dungeon[Next.x][Next.y][Next.z] != '#') {
                         vis[Next.x][Next.y][Next.z] = true;
                         Q.push(Next);
                     }
                 }
             }
         }
     }
     return -;
 }

 int main () {
     while(~scanf("%d %d %d", &l, &r, &c)) {
         if(l ==  && r ==  && c == )  break;
         for(int i = ; i < l; i ++) {
             for(int j = ; j < r; j ++) {
                 for(int k = ; k < c; k ++) {
                     cin >> dungeon[i][j][k];
                     if(dungeon[i][j][k] == 'S') { source.x = i; source.y = j; source.z = k; }
                     if(dungeon[i][j][k] == 'E') { key.x = i; key.y = j; key.z = k; }
                 }
             }
         }
         ans = bfs();
         if(ans != -) printf("Escaped in %d minute(s).\n", ans);
         else printf("Trapped!\n");
     }
     return ;
 }