Pushing Boxes (poj 1475 嵌套bfs)

Language:
Pushing Boxes
Time Limit: 2000MS Memory Limit: 131072K
Total Submissions: 4802 Accepted: 1653 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? 

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


题意:推箱子游戏,把箱子B推到T,人初始在S,输出推箱子次数最少的方案。一开始把题目看错,以为是总步数最小,写完交了果断WA,看了discuss才知道

思路:对箱子bfs,箱子移动一步再对人相应的bfs。实在无力吐槽,方向数组顺序必须是北南西东,调了一个下午,坑爹啊!!

代码:

#include <iostream>
#include <functional>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define DBG         pf("Hi\n")
typedef long long ll;
using namespace std;

#define INF 0x3f3f3f3f
#define mod 1000000009
const int maxn = 25;
const int MAXN = 2005;
const int MAXM = 2000100;
const int N = 1005;

struct Node
{
    int x,y;
    int Bx,By;
    string path;
}st,now;

int n,m,sx,sy,ex,ey,flag;
bool vis[maxn][maxn][maxn][maxn];
char mp[maxn][maxn];
int dir[4][2]={-1,0,1,0,0,-1,0,1};  //WA的检查一下方向数组,太坑爹
char a[]="nswe";
char A[]="NSWE";


bool isok(int x,int y)
{
    if (x>=1&&x<=n&&y>=1&&y<=m&&mp[x][y]!='#') return true;
    return false;
}

string bfsman(Node st,int Bx,int By,int tx,int ty,int &f)
{
    f=0;
//    printf("%d %d = %d %d = %d %d\n",Bx,By,sx,sy,tx,ty);
    Node tt,no;
    tt.Bx=Bx;tt.By=By;tt.x=st.x;tt.y=st.y;tt.path="";
    queue<Node>Q;
    vis[Bx][By][sx][sy]=true;
    Q.push(tt);
    while (!Q.empty())
    {
        tt=Q.front();Q.pop();
        if (tt.x==tx&&tt.y==ty)
        {
            f=1;
//            cout<<tt.path<<endl;
            return tt.path;
        }
        for (int i=0;i<4;i++)
        {
            no.x=tt.x+dir[i][0];
            no.y=tt.y+dir[i][1];
            if (no.x==st.Bx&&no.y==st.By) continue;
            if (isok(no.x,no.y)&&!vis[Bx][By][no.x][no.y])
            {
//                DBG;
                vis[Bx][By][no.x][no.y]=true;
                no.path=tt.path+a[i];
                Q.push(no);
            }
        }
    }
    return "";
}

void bfsbox()
{
    st.path="";
    memset(vis,false,sizeof(vis));
    vis[st.Bx][st.By][st.x][st.y]=true;
    queue<Node>Q;
    Q.push(st);
    while (!Q.empty())
    {
        st=Q.front();Q.pop();
        if (st.Bx==ex&&st.By==ey)
        {
            flag=1;
            cout<<st.path<<endl;
            return ;
        }
        for (int i=0;i<4;i++)
        {
            now.Bx=st.Bx+dir[i][0];
            now.By=st.By+dir[i][1];
            if (isok(now.Bx,now.By)&&!vis[now.Bx][now.By][st.x][st.y])
            {
                int dx=st.Bx-dir[i][0];
                int dy=st.By-dir[i][1];
                if (isok(dx,dy))
                {
                    vis[now.Bx][now.By][st.x][st.y]=true;
                    int xxx;
                    string ss=bfsman(st,now.Bx,now.By,dx,dy,xxx);
                    if (xxx)
                    {
//                        cout<<"cao "<<i<<ss<<endl;
                        now.path=st.path+ss+A[i];
                        now.x=st.Bx;now.y=st.By;
                        Q.push(now);
                    }
                    else
                        vis[now.Bx][now.By][st.x][st.y]=false;
                }
            }
        }
    }
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("C:/Users/lyf/Desktop/IN.txt","r",stdin);
#endif
    int i,j,cas=0;
    while (scanf("%d%d",&n,&m))
    {
        if (n==0&&m==0) break;
        ex=ey=-1;
        for (i=1;i<=n;i++)
        {
            scanf("%s",mp[i]+1);
            for (j=1;j<=m;j++)
            {
                if (mp[i][j]=='T')
                {
                    mp[i][j]='.';
                    ex=i;
                    ey=j;
                }
                if (mp[i][j]=='S')
                {
                    mp[i][j]='.';
                    st.x=i;st.y=j;
                }
                if (mp[i][j]=='B')
                {
                    mp[i][j]='.';
                    st.Bx=i;st.By=j;
                }
            }
        }
        flag=0;
        printf("Maze #%d\n",++cas);
//        cout<<bfsman(4,4,4,6,4,6)<<endl;
        bfsbox();
        if (!flag) printf("Impossible.\n");
        printf("\n");
    }
    return 0;
}


### 解决 Git Push 一直处于卡住状态的方法 当遇到 `git push` 卡住的情况时,可能的原因有多种。以下是常见的原因及其解决方案: #### 1. 网络连接问题 网络不稳定可能导致推送过程被中断或延迟。可以尝试以下方法来验证并解决问题: - 使用 `-v` 参数查看详细的日志信息以便定位具体问题[^1]。 ```bash git push -v ``` 如果发现请求超时或者无法建立连接,则可能是网络配置有问题。 #### 2. 远程仓库地址错误 确认远程仓库URL是否正确设置。可以通过运行命令检查当前配置的远程仓库地址[^2]: ```bash git remote -v ``` 如果有误, 可重新设定正确的SSH/HTTPS URL: ```bash git remote set-url origin git@github.com:username/repo.git ``` #### 3. 大文件上传限制 某些平台(如GitHub,GitLab)对于单次提交中的大文件大小有限制。如果存在超出规定尺寸的大文件,可能会导致传输失败。建议安装并启用Git LFS (Large File Storage)[^3], 它专门用于管理大型二进制数据集。 先初始化LFS跟踪特定类型的文件后再执行常规操作即可改善状况: ```bash git lfs install git lfs track "*.psd" git add .gitattributes git add yourfile.psd git commit -m "Add file with LFS support" git push ``` #### 4. 身份认证失效 有时由于令牌过期或者其他安全策略调整造成身份校验不通过也会引发此类现象。此时需更新个人访问token 或者重新生成ssh key 并将其添加到对应的服务器上完成绑定流程[^4]. #### 5. 防火墙代理干扰 企业内部环境下的防火墙规则有可能阻止对外部资源发起的数据交换活动。联系管理员了解具体情况并将必要的端口开放出来有助于恢复正常功能表现[^5]. 以上就是针对“Git一直处于Pushing状态”的一些排查思路和技术手段介绍。
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