Joe works in a maze. Unfortunately, portions of the maze have
caught on re, and the owner of the maze neglected to create a re
escape plan. Help Joe escape the maze.
Given Joe's location in the maze and which squares of the maze
are on re, you must determine whether Joe can exit the maze before
the re reaches him, and how fast he can do it.
Joe and the re each move one square per minute, vertically or
horizontally (not diagonally). The re spreads all four directions
from each square that is on re. Joe may exit the maze from any
square that borders the edge of the maze. Neither Joe nor the re
may enter a square that is occupied by a wall.
Input
The rst line of input contains a single integer, the number of test
cases to follow. The rst line of each test case contains the two
integers R and C, separated by spaces, with 1  R;C  1000. The
following R lines of the test case each contain one row of the maze. Each of these lines contains exactly
C characters, and each of these characters is one of:
 #, a wall
 ., a passable square
 J, Joe's initial position in the maze, which is a passable square
 F, a square that is on re
There will be exactly one J in each test case.
Output
For each test case, output a single line containing `IMPOSSIBLE' if Joe cannot exit the maze before the
re reaches him, or an integer giving the earliest time Joe can safely exit the maze, in minutes.
Sample Input
2
4 4
####
#JF#
#..#
#..#
3 3
###
#J.
#.F
Sample Output
3
IMPOSSIBLE

題目大意：Joe在一個迷宮裡工作，然後迷宮著火了，火擴散的速度和Joe移動的速度一樣都是每分鐘一格，'#'代表著牆，而Joe和火都只能走'.'所代表的廣場，問Joe能否在火燒到他之前逃出迷宮。

思路兩次bfs：將火蔓延的時間進行預處理，用一次BFS搜尋出火到達每一個廣場所需要的最短時間，然後再BFS人到各個廣場所需要的時間，如果人到某一個地點的時間少於火到達這裡的這時間，或者人到達的某一個地點，火燒不到，就可以繼續搜尋，直到Joe到矩陣的邊界為止

  1 //
  2 // Created by w on 2020/11/13.
  3 //
  4 
  5 #include<iostream>
  6
 
 #include<cstdio>
  7 #include<queue>
  8 #include<cstring>
  9 using namespace std;
 10 const int maxn=1010;
 11 const int inf=0x3f3f3f3f;
 12 typedef pair<int,int> P;
 13 int sx,sy,cnt,R,C,T,final;
 14 int s[maxn][maxn];
 15 int judge[maxn][maxn];
 16 char map[maxn][maxn];
 17 int
 
 dx[]={0,1,-1,0};
 18 int dy[]={1,0,0,-1};
 19 
 20 bool judgein1(int nx,int ny)
 21 {
 22     if(nx>=0&&ny>=0&&ny<C&&nx<R&&map[nx][ny]=='.'&&s[nx][ny]==inf)
 23         return true;
 24     return false;
 25 }
 26 bool judgein2(int nx,int ny)
 27 {
 28     if(nx>=0 && ny>=0 && ny<C &&nx<R&&map[nx][ny]!='#'&&judge[nx][ny]==inf)
 29         return true;
 30     return false;
 31 }
 32 void Joe_bfs()
 33 {
 34     queue<P >que;
 35     que.push(P(sx,sy));//存放初始位置座標pair(sx,sy)
 36     s[sx][sy]=0;
 37     while(que.size())
 38     {
 39         P p=que.front();
 40         que.pop();
 41         for(int i=0;i<4;i++)//遍歷各個方向
 42         {
 43             int nx=p.first+dx[i];
 44             int ny=p.second+dy[i];
 45             if(judgein1(nx,ny))
 46             {
 47                 que.push(P(nx,ny));
 48                 s[nx][ny]=s[p.first][p.second]+1;//記錄火燃燒的時間，用每一個新的座標存放時間
 49             }
 50         }
 51     }
 52 }
 53 void Fire_bfs()
 54 {
 55     memset(judge,inf,sizeof(judge));//重新初始化judge陣列
 56     queue<P>que;
 57     for(int i=0;i<R;i++)
 58     {
 59         for(int j=0;j<C;j++)
 60         {
 61             if(map[i][j]=='F')
 62             {
 63                 que.push(P(i,j));//入隊
 64                 judge[i][j]=0;//標記火的位置
 65             }
 66         }
 67     }
 68     while(que.size())
 69     {
 70         P p=que.front();
 71         que.pop();
 72         for(int i=0;i<4;i++)
 73         {
 74             int nx=p.first+dx[i];
 75             int ny=p.second+dy[i];
 76             if(judgein2(nx,ny))
 77             {
 78                 judge[nx][ny]=judge[p.first][p.second]+1;//記錄時間
 79                 if(judge[nx][ny]<=s[nx][ny])//如果被火燒到了，此位置設為-1
 80                     s[nx][ny]=-1;
 81                 que.push(P(nx,ny));
 82             }
 83         }
 84     }
 85 }
 86 int main()
 87 {
 88     ios::sync_with_stdio(false);
 89     cin>>T;
 90     while(T--)
 91     {
 92         final=inf;//初始化
 93         memset(s,inf,sizeof(s));
 94         memset(map,0,sizeof(map));
 95         cnt=0;
 96         cin>>R>>C;
 97         for(int i=0;i<R;i++)
 98         {
 99             for(int j=0;j<C;j++)
100             {
101                 cin>>map[i][j];
102                 if(map[i][j]=='J')//記錄初始位置
103                 {
104                     sx=i;
105                     sy=j;
106                 }
107             }
108         }
109         Joe_bfs();
110         Fire_bfs();
111         for(int i=0;i<R;i++)
112         {
113             for(int j=0;j<C;j++)
114             {
115                 if(i==R-1||j==C-1||i==0||j==0)
116                     if(s[i][j]!=-1&&s[i][j]!=inf)//這個點如果沒有被火燒到或者沒有對
117                     final=min(final,s[i][j]);
118             }
119         }
120         if(final!=inf)
121             cout<<final+1<<endl;
122         else
123             cout<<"IMPOSSIBLE"<<endl;
124     }
125     return 0;
126 }