POJ 2251 Dungeon Master(三维广搜——搜索方向（东,南,西,北,上,下）)

题目链接：https://cn.vjudge.net/problem/POJ-2251

题意：有l层平面，每个平面的大小是r*c,问从Ｓ到Ｅ最小的花费

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a));
using namespace std;
#define INF 0x3f3f3f3f
typedef long long ll;
int dir[4][2] = {0,1,0,-1,1,0,-1,0};
const int maxn = 50005;
int l,r,c,vis[40][40][40],ans;
char s[40][40][40];
bool flag;
struct node {
    int x,y,z,val;//val为起点到该点的花费
    node(int xx,int yy,int zz,int vall):x(xx),y(yy),z(zz),val(vall){};
};
void bfs(int ii,int jj,int kk) {//(z,x,y)
    queue<node>q;
    q.push(node(jj,kk,ii,0));//(x,y,z)
    vis[ii][jj][kk] = 1;
    while(!q.empty()) {
        node temp = q.front();
        q.pop();
        int fx = temp.x,fy=temp.y,fz =temp.z,fv = temp.val;
        if(s[fz][fx][fy] == 'E') {
            flag = 1;
            ans = fv;
            break;
        }
        for(int i = 0; i < 4; i++) {//东南西北四个方向
            fx = temp.x + dir[i][0],fy = temp.y + dir[i][1],fz = temp.z,fv = temp.val;
            if(fx >=0 && fx < r && fy >=0 && fy < c && (s[fz][fx][fy] == '.' || s[fz][fx][fy] == 'E') && !vis[fz][fx][fy])
            {
                vis[fz][fx][fy] = 1;
                q.push(node(fx,fy,fz,fv+1));//每走一步，花费加１
            }
        }
        fx = temp.x, fy = temp.y,fv = temp.val;
        fz = temp.z+1;
        if(fz >=0 && fz < l&&(s[fz][fx][fy] == '.' || s[fz][fx][fy] == 'E')&& !vis[fz][fx][fy]) {//向上
            vis[fz][fx][fy] = 1;
            q.push(node(fx,fy,fz,fv+1));
        }
        fz = temp.z-1;
        if(fz >=0 && fz < l&&(s[fz][fx][fy] == '.' || s[fz][fx][fy] == 'E') && !vis[fz][fx][fy]) {//向下
            vis[fz][fx][fy] = 1;
            q.push(node(fx,fy,fz,fv+1));
        }
    }
}
int main()
{

    while(cin >> l >> r >> c &&l&&r&&c) {
        mem(vis,0);
        flag = 0;
        for(int i = 0; i < l; i++) {
            for(int j = 0 ; j < r; j++) {
                cin >> s[i][j];
            }
        }
        for(int i = 0; i < l; i++) {
            for(int j = 0; j < r; j++) {
                for(int k = 0; k < c; k++) {
                    if(s[i][j][k] == 'S'){//找到起点开始搜
                        bfs(i,j,k);
                        break;
                    }
                }
            }
        }
        if(flag) cout << "Escaped in "<< ans << " minute(s)." <<endl;
        else cout << "Trapped!" << endl;
    }
    return 0;
}