You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes which may or may not be filled with rock. It takes one minute to move one unit north, south, east, west, up or down. You cannot move diagonally and the maze is surrounded by solid rock on all sides.
Is an escape possible? If yes, how long will it take?
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.
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!
3 4 5 S.... .###. .##.. ###.# ##### ##### ##.## ##... ##### ##### #.### ####E 1 3 3 S## #E# ### 0 0 0
Escaped in 11 minute(s). Trapped!
题目大意：三维空间，给定起点，给定终点，给定地图，求起点到终点的最短路。
#include<stdio.h>
#include<string.h>
int move[6][3] = {{-1, 0, 0}, {1, 0, 0}, {0, -1, 0}, {0, 1, 0}, {0, 0, -1}, {0, 0, 1}}; //上下前后左右
struct pos {
int x, y, z;
int time;
};
char map[40][40][40];
int vis[40][40][40], head, tail, min, flag;
pos p[1000000];
void BFS() {
int px, py, pz;
while (head < tail) {
if (map[p[head].x][p[head].y][p[head].z] == 'E') {
if (p[head].time < min) min = p[head].time;
flag = 1;
}
for (int i = 0; i < 6; i++) {
px = p[head].x + move[i][0];
py = p[head].y + move[i][1];
pz = p[head].z + move[i][2];
if (map[px][py][pz] != '#' && map[px][py][pz] != '@' && !vis[px][py][pz]) {
vis[px][py][pz] = 1;
p[tail].x = px;
p[tail].y = py;
p[tail].z = pz;
p[tail].time = p[head].time + 1;
tail++;
}
}
head++;
}
}
int main() {
int L, R, C;
while (scanf("%d %d %d", &L, &R, &C) == 3, L, R, C) {
memset(map, 0, sizeof(map));
memset(vis, 0, sizeof(vis));
for (int i = 1; i <= L; i++) {
for (int j = 1; j <= R; j++) {
getchar();
for (int k = 1; k <= C; k++) {
scanf("%c", &map[i][j][k]);
if (map[i][j][k] == 'S') {
p[1].x = i;
p[1].y = j;
p[1].z = k;
}
}
}
getchar();
}
for (int i = 0; i <= L + 1; i++) {
for (int j = 0; j <= R + 1; j++) {
map[i][j][0] = '@';
map[i][j][C + 1] = '@';
}
}
for (int i = 0; i <= L + 1; i++) {
for (int j = 0; j <= C + 1; j++) {
map[i][0][j] = '@';
map[i][R + 1][j] = '@';
}
}
for (int i = 0; i <= R + 1; i++) {
for (int j = 0; j <= C + 1; j++) {
map[0][i][j] = '@';
map[L + 1][i][j] = '@';
}
}
vis[p[1].x][p[1].y][p[1].z] = 1;
head = 1;
tail = 2;
flag = 0;
min = 9999;
p[1].time = 0;
BFS();
if (flag) printf("Escaped in %d minute(s).\n", min);
else printf("Trapped!\n");
}
return 0;
}
原文地址：http://blog.csdn.net/llx523113241/article/details/43407291
