uva 532(bfs)

题解：按照bfs的套路，不过多了一个三维数组，先是z轴，然后是x、y轴。

#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;
const int N = 35;

struct P {
	int x, y, z;
}p;

int l, r, c, x1, y1, z1, x2, y2, z2, flag1;
char pos[N][N][N];
int vis[N][N][N];
int flag[6][3] = {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}, {-1, 0, 0}, {0, -1, 0}, {0, 0, -1}};

queue<P> q;

void init() {
	for (int i = 0; i < N; i++)
        for (int j = 0; j < N; j++)
            for (int k = 0; k < N; k++) {
                pos[i][j][k] = '\0';
                vis[i][j][k] = 0;
            }
	flag1 = 0;
	while (!q.empty())
		q.pop();
}

void bfs() {
	int x0, y0, z0;
	while (!q.empty()) {
		for (int i = 0; i < 6; i++) {
			x0 = x1 + flag[i][0];
			y0 = y1 + flag[i][1];
			z0 = z1 + flag[i][2];
			if (x0 < 0 || x0 >= r || y0 < 0 || y0 >= c || z0 < 0 || z0 >= l)
				continue;
			if (vis[z0][x0][y0] == 0 && (pos[z0][x0][y0] == '.' || pos[z0][x0][y0] == 'E')) {
				vis[z0][x0][y0] = vis[z1][x1][y1] + 1;
				if (pos[z0][x0][y0] == 'E') {
                    flag1 = 1;
					return;
				}
				p.x = x0;
				p.y = y0;
				p.z = z0;
				q.push(p);
			}
		}
		q.pop();
		p = q.front();
		x1 = p.x;
		y1 = p.y;
		z1 = p.z;
	}
}

void found() {
	for (int i = 0; i < l; i++)
		for (int j = 0; j < r; j++)
			for (int k = 0; k < c; k++) {
				if (pos[i][j][k] == 'S') {
					x1 = p.z = j;
					y1 = p.x = k;
					z1 = p.y = i;
					q.push(p);
					bfs();
					continue;
				}
				if (pos[i][j][k] == 'E') {
					x2 = j;
					y2 = k;
					z2 = i;
					continue;
				}
			}
}
int main() {
	while (scanf("%d%d%d", &l, &r, &c) && l) {
        getchar();
        init();
		for (int i = 0; i < l; i++) {
			for (int j = 0; j < r; j++)
                scanf("%s", pos[i][j]);
            getchar();
		}
		getchar();
		found();
		if (flag1 == 0)
			printf("Trapped!\n");
		else
			printf("Escaped in %d minute(s).\n", vis[z2][x2][y2]);
        init();
	}
	return 0;
}