Pushing Boxes

Time Limit: 2000ms
Memory Limit: 131072KB

This problem will be judged on PKU. Original ID: 1475
64-bit integer IO format: %lld      Java class name: Main

Special Judge
 
Imagine you are standing inside a two-dimensional maze composed of square cells which may or may not be filled with rock. You can move north, south, east or west one cell at a step. These moves are called walks. 
One of the empty cells contains a box which can be moved to an adjacent free cell by standing next to the box and then moving in the direction of the box. Such a move is called a push. The box cannot be moved in any other way than by pushing, which means that if you push it into a corner you can never get it out of the corner again.

One of the empty cells is marked as the target cell. Your job is to bring the box to the target cell by a sequence of walks and pushes. As the box is very heavy, you would like to minimize the number of pushes. Can you write a program that will work out the best such sequence? 

 

Input

The input contains the descriptions of several mazes. Each maze description starts with a line containing two integers r and c (both <= 20) representing the number of rows and columns of the maze.

Following this are r lines each containing c characters. Each character describes one cell of the maze. A cell full of rock is indicated by a `#' and an empty cell is represented by a `.'. Your starting position is symbolized by `S', the starting position of the box by `B' and the target cell by `T'.

Input is terminated by two zeroes for r and c.

 

Output

For each maze in the input, first print the number of the maze, as shown in the sample output. Then, if it is impossible to bring the box to the target cell, print ``Impossible.''.

Otherwise, output a sequence that minimizes the number of pushes. If there is more than one such sequence, choose the one that minimizes the number of total moves (walks and pushes). If there is still more than one such sequence, any one is acceptable.

Print the sequence as a string of the characters N, S, E, W, n, s, e and w where uppercase letters stand for pushes, lowercase letters stand for walks and the different letters stand for the directions north, south, east and west.

Output a single blank line after each test case.

 

Sample Input

1 7
SB....T
1 7
SB..#.T
7 11
###########
#T##......#
#.#.#..####
#....B....#
#.######..#
#.....S...#
###########
8 4
....
.##.
.#..
.#..
.#.B
.##S
....
###T
0 0

Sample Output

Maze #1
EEEEE Maze #2
Impossible. Maze #3
eennwwWWWWeeeeeesswwwwwwwnNN Maze #4
swwwnnnnnneeesssSSS

Source

 
解题:二维bfs?。。。。。推箱子,主要方向是箱子移动方向,先确定箱子移动方向,再确定人要到达箱子的哪一侧,箱子移动,需要一个bfs,人到箱子的一侧需要一个bfs。。。故需要两个bfs。
 
 #include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = ;
struct stats{
int px,py,bx,by;
string path;
};
struct node{
int x,y;
string path;
};
char mp[maxn][maxn];
int r,c,box_x,box_y,pson_x,pson_y;
string ans;
const int dir[][] = {,-,,,-,,,};
const char dc[] = {'W','E','N','S'};
const char dc2[] = {'w','e','n','s'};
bool check(int x,int y){
return mp[x][y] != '#';
}
bool bfs2(int nx,int ny,int tx,int ty,int kx,int ky,string &pans){
queue<node>q;
bool vis[maxn][maxn] = {false};
vis[nx][ny] = vis[kx][ky] = true;
node now,tmp;
now.x = nx;
now.y = ny;
now.path = "";
q.push(now);
while(!q.empty()){
now = q.front();
q.pop();
if(now.x == tx && now.y == ty){
pans = now.path;
return true;
}
for(int i = ; i < ; i++){
int zx = now.x + dir[i][];
int zy = now.y + dir[i][];
if(check(zx,zy)&&!vis[zx][zy]){
vis[zx][zy] = true;
tmp.x = zx;
tmp.y = zy;
tmp.path = now.path + dc2[i];
q.push(tmp);
}
}
}
return false;
}
bool bfs(){
queue<stats>q;
bool vis[maxn][maxn] = {false};
vis[box_x][box_y] = true;
stats tmp,now;
now.px = pson_x;
now.py = pson_y;
now.bx = box_x;
now.by = box_y;
now.path = "";
q.push(now);
while(!q.empty()){
now = q.front();
q.pop();
for(int i = ; i < ; i++){
int nx = now.bx + dir[i][];
int ny = now.by + dir[i][];
int tx = now.bx - dir[i][];
int ty = now.by - dir[i][];
string pans = "";
if(check(nx,ny)&&check(tx,ty)&&!vis[nx][ny]){
if(bfs2(now.px,now.py,tx,ty,now.bx,now.by,pans)){
vis[nx][ny] = true;
tmp.px = now.bx;
tmp.py = now.by;
tmp.bx = nx;
tmp.by = ny;
tmp.path = now.path + pans + dc[i];
if(mp[nx][ny] == 'T'){
ans = tmp.path;
return true;
}
q.push(tmp);
}
}
}
}
return false;
}
int main(){
int cs = ;
while(~scanf("%d %d",&r,&c) && r + c){
memset(mp,'#',sizeof(mp));
getchar();
for(int i = ; i <= r; i++){
for(int j = ; j <= c; j++){
mp[i][j] = getchar();
if(mp[i][j] == 'B'){
box_x = i;
box_y = j;
}
if(mp[i][j] == 'S'){
pson_x = i;
pson_y = j;
}
}
getchar();
}
printf("Maze #%d\n", cs++);
if(bfs()) cout<<ans<<endl;
else puts("Impossible.");
puts("");
}
return ;
}

HDU 1475 Pushing Boxes的更多相关文章

  1. (poj 1475) Pushing Boxes

    Imagine you are standing inside a two-dimensional maze composed of square cells which may or may not ...

  2. POJ 1475 Pushing Boxes 搜索- 两重BFS

    题目地址: http://poj.org/problem?id=1475 两重BFS就行了,第一重是搜索箱子,第二重搜索人能不能到达推箱子的地方. AC代码: #include <iostrea ...

  3. poj 1475 Pushing Boxes 推箱子(双bfs)

    题目链接:http://poj.org/problem?id=1475 一组测试数据: 7 3 ### .T. .S. #B# ... ... ... 结果: //解题思路:先判断盒子的四周是不是有空 ...

  4. poj 1475 || zoj 249 Pushing Boxes

    http://poj.org/problem?id=1475 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=249 Pushin ...

  5. 『Pushing Boxes 双重bfs』

    Pushing Boxes Description Imagine you are standing inside a two-dimensional maze composed of square ...

  6. [poj P1475] Pushing Boxes

    [poj P1475] Pushing Boxes Time Limit: 2000MS   Memory Limit: 131072K   Special Judge Description Ima ...

  7. POJ1475 Pushing Boxes(双搜索)

    POJ1475 Pushing Boxes  推箱子,#表示墙,B表示箱子的起点,T表示箱子的目标位置,S表示人的起点 本题没有 Special Judge,多解时,先最小化箱子被推动的次数,再最小化 ...

  8. Pushing Boxes(广度优先搜索)

    题目传送门 首先说明我这个代码和lyd的有点不同:可能更加复杂 既然要求以箱子步数为第一关键字,人的步数为第二关键字,那么我们可以想先找到箱子的最短路径.但单单找到箱子的最短路肯定不行啊,因为有时候不 ...

  9. Pushing Boxes POJ - 1475 (嵌套bfs)

    Imagine you are standing inside a two-dimensional maze composed of square cells which may or may not ...

随机推荐

  1. 2014年辛星解读css第六节

    这一节我们就要讲到布局了,事实上布局本身特别简单.可是要合理的布好局就不那么简单了,就像我们写文章一样.写一篇文章非常easy,可是要写一篇名著就非常难了,这须要我们扎实的功底和对文学的理解,可是.千 ...

  2. 数据结构(三)——栈Stack

    栈是一种特殊的线性表,插入和删除操作均在栈顶进行,插入操作称为入栈,删除操作称为出栈. 一.顺序栈 利用顺序存储方式实现的栈称为顺序栈,下面是它的一些基本操作实现算法,需要理解和记忆. 1.顺序栈的类 ...

  3. Web前端开发实战6:CSS实现导航菜单结合二级下拉式菜单的简单变换

    前面几篇博文都在讲导航菜单和二级下拉式菜单,事实上有非常多方法都能够实现的.详细的情况还要视情况而定. 在后面学习到jQuery框架之后,会有更丰富的动画效果.因为在学习Ajax和jQuery的初步阶 ...

  4. [HTML] 微信wap开发---页面自适应大小

    <meta name="viewport" content="width=device-width, initial-scale=0.5, minimum-scal ...

  5. 常见Java集合的实现细节

    1. Set和Map Set代表一种集合元素无序.集合元素不可重复的集合,Map则代表一种由多个key-value对组成的集合,Map集合类似于传统的关联数组.表面上看它们之间相似性很少,但实际上Ma ...

  6. php pdo操作

    PDO(PHP Data Object) 是PHP 5 中加入的东西,是PHP 5新加入的一个重大功能,因为在PHP 5以前的php4/php3都是一堆的数据库扩展来跟各个数据库的连接和处理,什么 p ...

  7. 一篇个人感觉比较好的lua入门的文章

    原文转自www.cppprog.com,由三篇文章组成 Lua是一个嵌入式的脚本语言,它不仅可以单独使用还能与其它语言混合调用.Lua与其它脚本语言相比,其突出优势在于: 1.  可扩展性.Lua的扩 ...

  8. Hadoop MapReduce编程 API入门系列之join(二十六)(未完)

    不多说,直接上代码. 天气记录数据库 Station ID Timestamp Temperature 气象站数据库 Station ID Station Name 气象站和天气记录合并之后的示意图如 ...

  9. Solr.NET快速入门(七)【覆盖默认映射器,NHibernate集成】

    覆盖默认映射器 默认情况下,SolrNet使用属性映射Solr字段. 但是,您可能需要使用另一个映射程序. 替换默认映射器取决于您如何设置库: 内置容器 如果使用默认的内置容器,可以在调用Startu ...

  10. Solr快速入门(一)

    概述 本文档介绍了如何获取和运行Solr,将各种数据源收集到多个集合中,以及了解Solr管理和搜索界面. 首先解压缩Solr版本并将工作目录更改为安装Solr的子目录.请注意,基本目录名称可能随Sol ...