kuangbin专题一 Dungeon Master(BFS)

三维bfs，道理和二维是一样的

代码：

// zyc 2018/8/20

#include 
using namespace std;

const int maxn = 1e5 + 7;

char mp [100][100][100];
bool vis [100][100][100];
int fx [][3] = {0, 0, 1, 0, 0, -1, 0, 1, 0, 0, -1, 0, 1, 0, 0, -1, 0, 0};
int l, r, c;
struct node {
    int x, y, z;
    int step;
} ;
node s, e;
bool cheak (int i, int j, int k)
{
    if (mp[i][j][k] != '#' && (!vis[i][j][k]) && i >= 0 && i < l && j >= 0 && j < r && k >= 0 && k < c)
        return true;
    return false;
}
void bfs ()
{
    node s1, s2;
    vis [s.x][s.y][s.z] = true;
    queue  res;
    res.push(s);
    while (!res.empty()) {
        s1 = res.front ();
        res.pop();
        if (s1.x == e.x && s1.y == e.y && s1.z == e.z) {
            printf ("Escaped in %d minute(s).\n", s1.step);
            return ;
        }
        for (int i = 0; i < 6; i ++) {
            s2 = s1;
            s2.x += fx[i][0];
            s2.y += fx[i][1];
            s2.z += fx[i][2];
            if (cheak (s2.x, s2.y, s2.z)) {
                vis [s2.x][s2.y][s2.z] = true;
                s2.step ++;
                res.push(s2);
            }
        }
    }
    puts ("Trapped!");
}
int main ()
{
    while (~scanf ("%d %d %d", &l, &r, &c)) {
        if (l == 0 && r == 0 && c == 0) return 0;
        memset (vis, false, sizeof (vis));
        for (int i = 0; i < l; i ++) {
            for (int j = 0; j < r; j ++) {
                scanf (" %s", mp[i][j]);
                for (int k = 0; k < c; k ++) {
                    if (mp[i][j][k] == 'S') {
                        s.x = i, s.y = j, s.z = k;
                        s.step = 0;
                    } else if (mp[i][j][k] == 'E') {
                        e.x = i, e.y = j, e.z = k;
                    }
                }
            }
        }
        bfs ();
    }
    return 0;
}