poj 1475 Pushing Boxes

时间：2015-04-23 23:21:23 阅读：242 评论：0 收藏：0 [点我收藏+]

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http://poj.org/problem?id=1475

Pushing Boxes

Time Limit: 2000MS		Memory Limit: 131072K
Total Submissions: 4662		Accepted: 1608		Special Judge

Description

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?
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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

分析：
第一次看到题目的时候，以为可以两次bfs，先拿箱子bfs目标，记录路径，得到箱子的开始状态（即人的最终状态），然后再次拿人BFS箱子的初试状态，记录路径，把两个路径加起来即可（之前竟然不知道这个叫嵌套BFS）。
后来第三组数据不对，想到了箱子是不能像人一样拐弯的。

WA代码：

  1 #include <stdio.h>
  2 #include <algorithm>
  3 #include <iostream>
  4 #include <string.h>
  5 #include <string>
  6 #include <math.h>
  7 #include <stdlib.h>
  8 #include <queue>
  9 #include <stack>
 10 #include <set>
 11 #include <map>
 12 #include <list>
 13 #include <iomanip>
 14 #include <vector>
 15 #pragma comment(linker, "/STACK:1024000000,1024000000")
 16 #pragma warning(disable:4786)
 17 
 18 using namespace std;
 19 
 20 const int INF = 0x3f3f3f3f;
 21 const int MAX = 20 + 2;
 22 const double eps = 1e-8;
 23 const double PI = acos(-1.0);
 24 
 25 char ma[MAX][MAX];
 26 int vis[MAX][MAX];
 27 int n , m;
 28 int si , sj , bi , bj , ti , tj , ei , ej;
 29 int dir[4][2] = {1 , 0 , -1 , 0 , 0 , 1 , 0 , -1};
 30 char path[4] = {‘S‘ , ‘N‘ , ‘E‘ , ‘W‘};
 31 char path1[4] = {‘s‘ , ‘n‘ , ‘e‘ , ‘w‘};
 32 string str , str1 , ans;
 33 
 34 struct T
 35 {
 36     int x , y ;
 37     string sss;
 38 }temp , in , out;
 39 
 40 queue<T>qq , qq1;
 41 
 42 void bfs(int i , int j)
 43 {
 44     temp.x = i;
 45     temp.y = j;
 46     temp.sss = "";
 47     qq.push(temp);
 48     vis[i][j] = 1;
 49     while(!qq.empty())
 50     {
 51         out = qq.front();
 52         qq.pop();
 53         if(out.x == ti && out.y == tj)
 54         {
 55             str = out.sss;
 56             break;
 57         }
 58         for(int k = 0;k < 4;k ++)
 59         {
 60             int ix = out.x + dir[k][0];
 61             int iy = out.y + dir[k][1];
 62             if(ma[ix][iy] == ‘#‘ || vis[ix][iy] || ix < 1 || iy < 1 || ix > n || iy > m)
 63                 continue;
 64             in.x = ix;
 65             in.y = iy;
 66             in.sss = out.sss + path[k];
 67             vis[ix][iy] = 1;
 68             qq.push(in);
 69         }
 70     }
 71 }
 72 
 73 void bfs1(int i , int j)
 74 {
 75     temp.x = i;
 76     temp.y = j;
 77     temp.sss = "";
 78     qq.push(temp);
 79     vis[i][j] = 1;
 80     while(!qq.empty())
 81     {
 82         out = qq.front();
 83         qq.pop();
 84         if(out.x == ei && out.y == ej)
 85         {
 86             str1 = out.sss;
 87             break;
 88         }
 89         for(int k = 0;k < 4;k ++)
 90         {
 91             int ix = out.x + dir[k][0];
 92             int iy = out.y + dir[k][1];
 93             if(ma[ix][iy] == ‘#‘ || ma[ix][iy] == ‘B‘ || vis[ix][iy] || ix < 1 || iy < 1 || ix > n || iy > m)
 94                 continue;
 95             in.x = ix;
 96             in.y = iy;
 97             in.sss = out.sss + path1[k];
 98             vis[ix][iy] = 1;
 99             qq.push(in);
100         }
101     }
102 }
103 
104 int main()
105 {
106     int first = 1;
107     while(scanf("%d %d",&n , &m) , n + m)
108     {
109         int i , j;
110         memset(vis , 0 , sizeof(vis));
111         for(i = 1;i <= n;i ++)
112         {
113             for(j = 1;j <= m;j ++)
114             {
115                 scanf("%c",&ma[i][j]);
116                 if(ma[i][j] == ‘S‘)
117                 {
118                     si = i;
119                     sj = j;
120                 }
121                 else if(ma[i][j] == ‘B‘)
122                 {
123                     bi = i;
124                     bj = j;
125                 }
126                 else if(ma[i][j] == ‘T‘)
127                 {
128                     ti = i;
129                     tj = j;
130                 }
131             }
132             getchar();
133         }
134         str = "";
135         while(!qq.empty())
136             qq.pop();
137         bfs(bi , bj);
138         if(str[0] == ‘S‘)
139         {
140             ei = bi - 1;
141             ej = bj;
142         }
143         else if(str[0] == ‘N‘)
144         {
145             ei = bi + 1;
146             ej = bj;
147         }
148         else if(str[0] == ‘E‘)
149         {
150             ei = bi;
151             ej = bj - 1;
152         }
153         else if(str[0] == ‘W‘)
154         {
155             ei = bi;
156             ej = bj + 1;
157         }
158 
159         memset(vis , 0 , sizeof(vis));
160         str1 = "";
161         while(!qq.empty())
162             qq.pop();
163         bfs1(si , sj);
164         ans = "";
165         ans = str1 + str;
166 
167         printf("Maze #%d\n",first ++);
168         if(ans != "")
169             cout << ans << "\n" << endl;
170         else
171             cout << "Impossible\n" << endl;
172     }
173     return 0;
174 }

View Code

后来搜了一些资料。（明天再搞，，，）

AC代码：

  1 #include <iostream>
  2 #include <cstring>
  3 #include <cstdio>
  4 #include <algorithm>
  5 #include <queue>
  6 #include <string>
  7 
  8 using namespace std;
  9 
 10 #define MAXN 22
 11 char P[4]={‘N‘,‘S‘,‘W‘,‘E‘};
 12 char M[4]={‘n‘,‘s‘,‘w‘,‘e‘};
 13 int R,C;
 14 int dir[4][2]={-1,0,1,0,0,-1,0,1};
 15 char map[MAXN][MAXN];
 16 struct point
 17 {
 18     int x,y;
 19     int p_x,p_y;//当前状态person所在的地方
 20     string ans;
 21 };
 22 bool isok(int x,int y)
 23 {
 24     if(x>=0 && x<R && y>=0 && y<C && map[x][y] != ‘#‘)
 25         return true;
 26     return false;
 27 }
 28 string tmp;
 29 bool bfs_person(point en,point cu)
 30 {
 31     tmp="";
 32     point st;
 33     st.x=en.p_x;
 34     st.y=en.p_y;
 35     st.ans="";
 36     queue<point>q;
 37     bool vis[MAXN][MAXN];
 38     memset(vis,0,sizeof(vis));
 39     while(!q.empty())
 40         q.pop();
 41     q.push(st);
 42     while(!q.empty())
 43     {
 44         point cur,next;
 45         cur=q.front();
 46         q.pop();
 47         if(cur.x==en.x && cur.y==en.y)
 48         {
 49             tmp=cur.ans;
 50             return true;
 51         }
 52         for(int i=0;i<4;i++)
 53         {
 54             next=cur;
 55             next.x=cur.x+dir[i][0];
 56             next.y=cur.y+dir[i][1];
 57             if(!isok(next.x,next.y)) continue;
 58             if(next.x==cu.x && next.y==cu.y) continue;
 59             if(vis[next.x][next.y]) continue;
 60             vis[next.x][next.y]=1;
 61             next.ans=cur.ans+M[i];
 62             q.push(next);
 63         }
 64     }
 65     return false;
 66 }
 67 string bfs_box()
 68 {
 69     bool vis[MAXN][MAXN][4];//某点四个方向是否访问！！0==N,1==S,2==W,3==E
 70     point st;
 71     st.x=st.y=-1;
 72     st.p_x=st.p_y=-1;
 73     st.ans="";
 74     for(int i=0;i<R && (st.x==-1 || st.p_x==-1);i++)
 75         for(int j=0;j<C && (st.x==-1 || st.p_x==-1);j++)
 76             if(map[i][j]==‘B‘)
 77             {
 78                 st.x=i;
 79                 st.y=j;
 80                 map[i][j]=‘.‘;
 81             }
 82             else if(map[i][j]==‘S‘)
 83             {
 84                 st.p_x=i;
 85                 st.p_y=j;
 86                 map[i][j]=‘.‘;
 87             }
 88     //----------------------------------------
 89     //cout<<"st.x="<<st.x<<" st.y="<<st.y<<" st.p_x="<<st.p_x<<" st.p_y="<<st.p_y<<endl;
 90     //----------------------------------------
 91     queue<point> q;
 92     while(!q.empty())
 93         q.pop();
 94     q.push(st);
 95     memset(vis,0,sizeof(vis));
 96     while(!q.empty())
 97     {
 98         point cur=q.front();q.pop();    
 99         //----------------------------------------
100     //        cout<<"cur.x="<<cur.x<<" cur.y="<<cur.y<<" cur.p_x="<<cur.p_x<<" cur.p_y="<<cur.p_y<<endl;
101     //    cout<<"-----------------------------\n";
102         //----------------------------------------
103         point next,pre;
104         if(map[cur.x][cur.y]==‘T‘)
105             return cur.ans;
106         for(int i=0;i<4;i++)
107         {
108             next=cur;
109             next.x=cur.x+dir[i][0];
110             next.y=cur.y+dir[i][1];
111             if(!isok(next.x,next.y))
112                 continue;
113             if(vis[next.x][next.y][i])
114                 continue;
115             pre=cur;
116             switch(i)
117             {
118                 case 0: pre.x=cur.x+1;break;
119                 case 1: pre.x=cur.x-1;break;
120                 case 2: pre.y=cur.y+1;break;
121                 case 3: pre.y=cur.y-1;break;
122             }
123             if(!bfs_person(pre,cur))//搜寻人是否能走到特定的位置
124                 continue;
125             vis[next.x][next.y][i]=1;
126             next.ans=cur.ans+tmp;
127             next.ans=next.ans+P[i];
128             next.p_x=cur.x;next.p_y=cur.y;
129             q.push(next);
130         }
131     }
132     return "Impossible.";
133 }
134 
135 int main()
136 {
137     int cas=1;
138     while(scanf("%d%d",&R,&C) && (R+C))
139     {
140         getchar();
141         for(int i=0;i<R;i++)
142             gets(map[i]);
143 
144         //---------------------------------------
145     //    for(int i=0;i<R;i++)
146         //    cout<<map[i]<<endl;
147         //----------------------------------------
148 
149         printf("Maze #%d\n",cas++);
150         //printf("%s\n",bfs_box());
151         cout<<bfs_box()<<endl<<endl;
152     }
153     return 0;
154 }

View Code

poj 1475 Pushing Boxes

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原文地址：http://www.cnblogs.com/jeff-wgc/p/4451932.html

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