题目：
1096. 地牢大师 – AcWing题库
输入样例：
3 4 5S.....###..##..###.#############.####...###########.#######E1 3 3S###E####0 0 0输出样例：
Escaped in 11 minute(s).Trapped!思路：bfs（三维）
代码：
#includeusing namespace std;const int N=110;char p[N][N][N];//三维空间某位置上的字符int dist[N][N][N];//距离起点的距离struct Point//坐标{int x,y,z;};int L,R,C;int bfs(Point Start,Point End){queueq;memset(dist,-1,sizeof dist);q.push(Start);dist[Start.x][Start.y][Start.z]=0;while(q.size()){Point t=q.front();q.pop();int dx[6]={1,-1,0,0,0,0},dy[6]={0,0,1,-1,0,0},dz[6]={0,0,0,0,1,-1};for(int i=0;i<6;i++){int x=t.x+dx[i],y=t.y+dy[i],z=t.z+dz[i];if(x = L || y = R || z = C)continue;if(p[x][y][z]=='#')continue;if(dist[x][y][z]!=-1)continue;dist[x][y][z]=dist[t.x][t.y][t.z]+1;if(x==End.x&&y==End.y&&z==End.z)return dist[x][y][z];q.push({x,y,z});}}return -1;}int main(){cin>>L>>R>>C;Point Start,End;while(scanf("%d%d%d",&L,&R,&C),L||R||C)//输入000停止{for(int i=0;i<L;i++){for(int j=0;j<R;j++){scanf("%s",&p[i][j]);//按行输入for(int k=0;k<C;k++){if(p[i][j][k]=='S')Start={i,j,k};else if(p[i][j][k]=='E')End={i,j,k};}}}int distance=bfs(Start,End);if(distance==-1)printf("Trapped!\n");else printf("Escaped in %d minute(s).\n",distance);}return 0;}