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

    Dungeon Master

    Time Limit: 1000ms
    Memory Limit: 65536KB
    This problem will be judged on PKU. Original ID: 2251
    64-bit integer IO format: %lld      Java class name: Main
    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!
    

    Source

     
    解题:搜索。。。
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <cmath>
     5 #include <cstdlib>
     6 #include <algorithm>
     7 #include <stack>
     8 #include <set>
     9 #include <map>
    10 #include <queue>
    11 #include <ctime>
    12 #define LL long long
    13 #define INF 0x3f3f3f3f
    14 #define pii pair<int,int>
    15 
    16 using namespace std;
    17 const int maxn = 40;
    18 char table[maxn][maxn][maxn];
    19 int L,R,C,sx,sy,sz,ex,ey,ez;
    20 bool vis[maxn][maxn][maxn];
    21 struct node {
    22     int x,y,z,step;
    23     node(int tx = 0,int ty = 0,int tz = 0,int tp = 0) {
    24         x = tx;
    25         y = ty;
    26         z = tz;
    27         step = tp;
    28     }
    29 };
    30 const int dir[6][3] = {
    31     {0,0,-1},//left
    32     {0,0,1},//right
    33     {0,-1,0},//up
    34     {0,1,0},//down
    35     {-1,0,0},//front
    36     {1,0,0}//back
    37 };
    38 bool isIn(const node &tmp) {
    39     int a = (tmp.x >= 0 && tmp.x < L);
    40     int b = (tmp.y >= 0 && tmp.y < R);
    41     int c = (tmp.z >= 0 && tmp.z < C);
    42     return a + b + c == 3;
    43 }
    44 queue<node>q;
    45 int bfs() {
    46     while(!q.empty()) q.pop();
    47     q.push(node(sx,sy,sz,0));
    48     vis[sx][sy][sz] = true;
    49     while(!q.empty()) {
    50         node now = q.front();
    51         q.pop();
    52         if(table[now.x][now.y][now.z] == 'E') return now.step;
    53         for(int i = 0; i < 6; ++i) {
    54             node tmp(now.x+dir[i][0],now.y+dir[i][1],now.z+dir[i][2],now.step+1);
    55             if(isIn(tmp)&&table[tmp.x][tmp.y][tmp.z] != '#' &&!vis[tmp.x][tmp.y][tmp.z]) {
    56                 vis[tmp.x][tmp.y][tmp.z] = true;
    57                 if(table[tmp.x][tmp.y][tmp.z] == 'E') return tmp.step;
    58                 q.push(tmp);
    59             }
    60         }
    61     }
    62     return -1;
    63 }
    64 int main() {
    65     while(scanf("%d %d %d",&L,&R,&C),L||R||C) {
    66         memset(table,'',sizeof(table));
    67         memset(vis,false,sizeof(vis));
    68         for(int i = 0; i < L; ++i) {
    69             for(int j = 0; j < R; ++j) {
    70                 scanf("%s",table[i][j]);
    71                 for(int k = 0; k < C; ++k)
    72                     if(table[i][j][k] == 'S') {
    73                         sx = i;
    74                         sy = j;
    75                         sz = k;
    76                     }
    77             }
    78         }
    79         int ans = bfs();
    80         if(~ans) printf("Escaped in %d minute(s).
    ",ans);
    81         else puts("Trapped!");
    82     }
    83     return 0;
    84 }
    View Code
  • 相关阅读:
    window.showModalDialog使用手册
    javascrīpt 对象的定义
    导航上用CSS标志当前页效果的实现
    数据库中的命名规则
    Oracle数学函数
    实现 asp 的服务器无刷新推技术
    UML建模工具比较
    Oracle日期函数
    神秘的 ORACLE DUAL
    Podcast Publisher——一个综合了多种入门知识的小Web应用
  • 原文地址:https://www.cnblogs.com/crackpotisback/p/4236289.html
Copyright © 2011-2022 走看看