nyoj3533D dungeon

3D dungeon

时间限制：1000 ms  |  内存限制：65535 KB

难度：2

输入

The input consists of a number of dungeons. Each dungeon description starts with a line containing three integers L, R and C (all limited to 30 in size). 
L is the number of levels making up the dungeon. 
R and C are the number of rows and columns making up the plan of each level. 
Then there will follow L blocks of R lines each containing C characters. Each character describes one cell of the dungeon. A cell full of rock is indicated by a '#' and empty cells are represented by a '.'. Your starting position is indicated by 'S' and the exit by the letter 'E'. There's a single blank line after each level. Input is terminated by three zeroes for L, R and C.

输出

Each maze generates one line of output. If it is possible to reach the exit, print a line of the form 
Escaped in x minute(s).

where x is replaced by the shortest time it takes to escape. 
If it is not possible to escape, print the line 
Trapped!

样例输入

3 4 5S.....###..##..###.#############.####...###########.#######E1 3 3S###E####0 0 0

样例输出

Escaped in 11 minute(s).Trapped!

题目大意是：以S为起点，问你能否逃出三维牢房（到达E即为逃出），能则输出最短时间，每次只能上下前后左右移动一个位置，移动一个次用时1.

BFS即可AC

AC代码:

#include <stdio.h>char map[31][31][31], mark[31][31][31];int i, j, k, L, R, C, Z, Y, X, b[30000][4];int next[6][3]= {{0,0,-1},{0,1,0},{0,0,1},{0,-1,0},{-1,0,0},{1,0,0}};int bfs() {k = 0, j = 1, b[0][3] = 0;int z,y,x;while(k < j) { z = b[k][0], y = b[k][1], x = b[k][2]; for(int p = 0; p < 6; p++) {Z = z+next[p][0];Y = y+next[p][1];X = x+next[p][2];if(Z>=0&&Y>=0&&X>=0&&Z<L&&Y<R&&X<C) {if(map[Z][Y][X]=='.'&&mark[Z][Y][X]==0) {mark[Z][Y][X] = 1;  b[j][0] = Z;  b[j][1] = Y;  b[j][2] = X;  b[j++][3] = b[k][3]+1;  } else if(map[Z][Y][X]=='E') return b[k][3]+1;} } k++;}return 0;}int main() {while(scanf("%d %d %d", &L, &R, &C) && (L||R||C)) {for(i = 0; i < L; i++) {for(j = 0; j < R; j++) {  for(k = 0; k < C; k++) {  scanf(" %c", &map[i][j][k]);  mark[i][j][k] = 0;  if(map[i][j][k] == 'S') {  b[0][0] = i;  b[0][1] = j;  b[0][2] = k;  mark[i][j][k] = 1;  }  }  }}i = bfs();if(i) printf("Escaped in %d minute(s).\n", i);else printf("Trapped!\n");}return 0;}