zoukankan      html  css  js  c++  java
  • poj 2251 Dungeon Master

    题目连接

    http://poj.org/problem?id=2251

    Dungeon Master

    Description

    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? 

    Input

    The input consists of a number of dungeons. Each dungeon description starts with a line containing three integers L, R and C (all limited to 30 in size). 
    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.

    Output

    Each maze generates one line of output. If it is possible to reach the exit, print a line of the form
      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!

    Sample Input

    3 4 5
    S....
    .###.
    .##..
    ###.#

    #####
    #####
    ##.##
    ##...

    #####
    #####
    #.###
    ####E

    1 3 3
    S##
    #E#
    ###

    0 0 0

    Sample Output

    Escaped in 11 minute(s).
    Trapped!

    三维迷宫。。

     1 #include<algorithm>
     2 #include<iostream>
     3 #include<cstdlib>
     4 #include<cstring>
     5 #include<cstdio>
     6 #include<vector>
     7 #include<queue>
     8 #include<map>
     9 using std::cin;
    10 using std::cout;
    11 using std::endl;
    12 using std::find;
    13 using std::sort;
    14 using std::map;
    15 using std::pair;
    16 using std::queue;
    17 using std::vector;
    18 using std::multimap;
    19 #define pb(e) push_back(e)
    20 #define sz(c) (int)(c).size()
    21 #define mp(a, b) make_pair(a, b)
    22 #define all(c) (c).begin(), (c).end()
    23 #define iter(c) decltype((c).begin())
    24 #define cls(arr,val) memset(arr,val,sizeof(arr))
    25 #define cpresent(c, e) (find(all(c), (e)) != (c).end())
    26 #define rep(i, n) for (int i = 0; i < (int)(n); i++)
    27 #define tr(c, i) for (iter(c) i = (c).begin(); i != (c).end(); ++i)
    28 const int N = 32;
    29 typedef unsigned long long ull;
    30 char G[N][N][N];
    31 bool vis[N][N][N];
    32 int L, R, C, Sx, Sy, Sz, Dx, Dy, Dz;
    33 const int dx[] = { -1, 1, 0, 0, 0, 0 };
    34 const int dy[] = { 0, 0, -1, 1, 0, 0 }, dz[] = { 0, 0, 0, 0, -1, 1 };
    35 struct Node {
    36     int x, y, z, s;
    37     Node() {}
    38     Node(int i, int j, int k, int l) :x(i), y(j), z(k), s(l) {}
    39 };
    40 int bfs() {
    41     cls(vis, false);
    42     queue<Node> q;
    43     q.push(Node(Sx, Sy, Sz, 0));
    44     vis[Sx][Sy][Sz] = true;
    45     while (!q.empty()) {
    46         Node t = q.front(); q.pop();
    47         if (t.x == Dx && t.y == Dy && t.z == Dz) return t.s;
    48         rep(i, 6) {
    49             int x = t.x + dx[i], y = t.y + dy[i], z = t.z + dz[i];
    50             if (x < 0 || x >= L || y < 0 || y >= R || z < 0 || y >= C) continue;
    51             if (vis[x][y][z] || G[x][y][z] == '#') continue;
    52             q.push(Node(x, y, z, t.s + 1));
    53             vis[x][y][z] = true;
    54         }
    55     }
    56     return 0;
    57 }
    58 int main() {
    59 #ifdef LOCAL
    60     freopen("in.txt", "r", stdin);
    61     freopen("out.txt", "w+", stdout);
    62 #endif
    63     while (~scanf("%d %d %d", &L, &R, &C) && L + R + C) {
    64         rep(i, L) {
    65             rep(j, R) {
    66                 scanf("%s", G[i][j]);
    67                 rep(k, C) {
    68                     if (G[i][j][k] == 'S') Sx = i, Sy = j, Sz = k;
    69                     if (G[i][j][k] == 'E') Dx = i, Dy = j, Dz = k;
    70                 }
    71             }
    72         }
    73         int ret = bfs();
    74         if (!ret) puts("Trapped!");
    75         else printf("Escaped in %d minute(s).
    ", ret);
    76      }
    77     return 0;
    78 }
    View Code
    By: GadyPu 博客地址:http://www.cnblogs.com/GadyPu/ 转载请说明
  • 相关阅读:
    怎样使用两行代码实现博客园打赏功能
    使用vue开发微信公众号下SPA站点的填坑之旅
    贝叶斯公式与最大后验估计(MAP)
    多元高斯分布(The Multivariate normal distribution)
    Jacobian矩阵、Hessian矩阵和Newton's method
    导数、方向导数与梯度
    解决只有单引号的Json格式转换成bean问题
    浅析Java中的final关键字
    观察者模式/ java实现附代码 /
    Java内存区域与内存溢出异常
  • 原文地址:https://www.cnblogs.com/GadyPu/p/4621883.html
Copyright © 2011-2022 走看看