题解

其实这道题很简单 : )
代码:

#include <bits/stdc++.h>
using namespace std;
int n, m, u, ans = -1, a_n, a_m, a_u, ans_n, ans_m, ans_u;
char a[110][110][110];
bool vis[110][110][110];
int dir[6][3] = {{-1, 0, 0}, {1, 0, 0}, {0, -1, 0}, {0, 1, 0}, {0, 0, 1}, {0, 0, -1}};

struct Node{
	int x, y, z, step;
};

bool check(int x, int y, int z){
	return z >= 1 && z <= u && x >= 1 && y >= 1 && x <= n && y <= m && a[x][y][z] != '#' && vis[x][y][z] == false;
}

void bfs(int x, int y, int z, int step){
	queue<Node> q;
	q.push({x, y, z, step});
	vis[x][y][z] == true;
	while(!q.empty()){
		Node t = q.front();
		if(t.x == ans_n && t.y == ans_m && t.z == ans_u){
			ans = t.step;
			return ; 
		}
		q.pop();
		for(int i = 0;i <= 5;i++){
			int nx = t.x + dir[i][0];
			int ny = t.y + dir[i][1];
            int nz = t.z + dir[i][2];
			if(check(nx, ny, nz)){
				q.push({nx, ny, nz, t.step + 1});
				vis[nx][ny][nz] = true;
			}
		}
	}
}

int main(){
	cin >> n >> m >> u;
	for(int i = 1;i <= n;i++){
		for(int j = 1;j <= m;j++){
			for(int k = 1;k <= u;k++){
                cin >> a[i][j][k];
                if(a[i][j][k] == 'E'){
                    ans_n = i;
                    ans_m = j;
                    ans_u = k;
                }
                if(a[i][j][k] == 'S'){
                    a_n = i;
                    a_m = j;
                    a_u = k;
                }
            }
		}
	}
	bfs(a_n, a_m, a_u, 0);
	if(ans == -1) cout << "Trapped!";
    else cout << "Escaped in " << ans << " minute(s).";
	return 0;
}