POJ1475【推箱子】(双bfs)

先对箱子bfs，箱子想要推到那个位置，我们人就必须移动到箱子的后面去推它，因此每次推箱子，需要再对人bfs来判断人能不能走到推箱子需要的地方以及最少的步数，

struct rec {
    int bx, by, px, py;//箱子的坐标 人的坐标
    string path;//保存路径
    rec(int _bx = 0, int _by = 0, int _px = 0, int _py = 0, string _path = "") :bx(_bx), by(_by), px(_px), py(_py), path(_path){}
};

int bbx = cur.bx + dx[i];
int bby = cur.by + dy[i];
int ppx = cur.bx - dx[i];
int ppy = cur.by - dy[i];

bbx，bby就是箱子要推到的位置，ppx，ppy就是人推箱子需要走到的地方，cur.px,cur.py是当前人所在的地方，我们需要再进行一次bfs来判断能不能从当前位置走到目标位置，并且保存路径

能走到目标位置我们就把新状态的箱子加入队列，继续对箱子bfs

#include<bits/stdc++.h>
using namespace std;
struct rec {
    int bx, by, px, py;
    string path;
    rec(int _bx = 0, int _by = 0, int _px = 0, int _py = 0, string _path = "") :bx(_bx), by(_by), px(_px), py(_py), path(_path){}
};
struct person{
    int px, py;
    string path;
    person(int _px = 0, int _py = 0, string _path = ""):px(_px), py(_py), path(_path){}
};
char p[5] = { 'n', 's', 'e', 'w' };
char b[5] = { 'N', 'S', 'E', 'W' };
int dx[4] = { -1, 1, 0, 0 }, dy[4] = { 0, 0, 1, -1 };
char maze[25][25];
int n,m;
rec st;
int cas;
void parse()
{
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            if (maze[i][j] == 'S')
            {
                st.px = i, st.py = j;
            }
            else if (maze[i][j] == 'B')
            {
                st.bx = i, st.by = j;
            }
        }
    }
}
bool valid(int x, int y)
{
    if (x < 0 || x >= n || y < 0 || y >= m) return false;
    if (maze[x][y] == '#') return false;
    return true;
}
bool bfs1(int sx,int sy,int gx,int gy,int bbx,int bby,string &path)//判断能否从当前人在的位置（sx，sy）走到推箱子所需的位置（gx，gy）； （bbx，bby）是当前箱子的位置
{
    int vis[25][25];
    memset(vis, 0, sizeof(vis));
    vis[sx][sy] = 1;
    vis[bbx][bby] = 1;
    queue<person> q;
    person f;
    f.px = sx, f.py = sy, f.path = "";
    q.push(f);
    while (!q.empty())
    {
        person cur = q.front();
        q.pop();
        if (cur.px == gx&&cur.py == gy)
        {
            path = cur.path;
            return true;
        }
        for (int i = 0; i < 4; i++)
        {
            int nx = cur.px + dx[i], ny = cur.py + dy[i];
            if (valid(nx, ny) && !vis[nx][ny])
            {
                vis[nx][ny] = 1;
                person nxt;
                nxt.px = nx;
                nxt.py = ny;
                nxt.path = cur.path + p[i];
                q.push(nxt);
            }
        }
    }
    return false;
}
string  bfs()
{
    string ans;
    int flag = 0;
    int vis[25][25];
    memset(vis, 0, sizeof(vis));
    queue<rec> que;
    vis[st.bx][st.by] = 1;
    st.path = "";
    que.push(st);
    while (!que.empty())
    {
        rec cur = que.front();
        que.pop();
        for (int i = 0; i < 4; i++)
        {
            int bbx = cur.bx + dx[i], bby = cur.by + dy[i];
            int ppx = cur.bx - dx[i], ppy = cur.by - dy[i];
            string path = "";
            if (valid(bbx, bby) && valid(ppx, ppy) && !vis[bbx][bby])
            {
                if (bfs1(cur.px, cur.py, ppx, ppy, cur.bx, cur.by, path))
                {
                    vis[bbx][bby] = 1;
                    rec nxt;
                    nxt.bx = bbx;
                    nxt.by = bby;
                    nxt.px = cur.bx;
                    nxt.py = cur.by;
                    nxt.path = cur.path + path + b[i];
                    if (maze[bbx][bby] == 'T') 
                    { 
                        if (!flag)
                            ans = nxt.path,flag=1;
                        else
                        {
                            if ((nxt.path).size() <  ans.size() )
                            {
                                ans = nxt.path;
                            }
                        }
                    }
                    que.push(nxt);
                }
            }
        }
    }
    if (flag) return ans;
    else
    {
        ans = "Impossible.";
        return ans;
    }
    
}
int main()
{
    while (cin >> n >> m&&!(n == 0 && m == 0))
    {
        for (int i = 0; i < n; i++)
            scanf("%s", maze[i]);
        parse();
        printf("Maze #%d\n", ++cas);
        string ans = bfs();
        cout << ans << endl << endl;
    }
    return 0;
}

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