算法：BFS

Joe works in a maze. Unfortunately, portions of the maze have

caught on fire, and the owner of the maze neglected to create a fire

escape plan. Help Joe escape the maze.

Given Joe’s location in the maze and which squares of the maze

are on fire, you must determine whether Joe can exit the maze before

the fire reaches him, and how fast he can do it.

Joe and the fire each move one square per minute, vertically or

horizontally (not diagonally). The fire spreads all four directions

from each square that is on fire. Joe may exit the maze from any

square that borders the edge of the maze. Neither Joe nor the fire

may enter a square that is occupied by a wall.

Input

The first line of input contains a single integer, the number of test

cases to follow. The first 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 fire

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

fire 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

注意开始有多个火堆；

代码：

#include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #define INF 100000000using namespace std;char ch[1005][1005];int b[1005][1005],c[1004][1005],qq,cnt,d[1005][2];int n,m,a[4][2]={-1,0,0,-1,0,1,1,0},p,q;struct dot{ int x,y,step; };void bfs(){ memset(c,0,sizeof(c)); queue
          
           que; dot cur,loer; for(int i=0;i
           
            =0&&dx
            
             =0&&dy
             
              que; dot cur,loer; cur.x=p; cur.y=q; cur.step=0; c[p][q]=1; que.push(cur); while(que.size()) { loer=que.front(); que.pop(); if(loer.x==0||loer.x==n-1||loer.y==0||loer.y==m-1) { qq=1; cout<
              
               <
               
                =0&&dx
                
                 =0&&dy
                 
                  loer.step+1) { cur.x=dx; cur.y=dy; cur.step=loer.step+1; c[dx][dy]=1; que.push(cur); } } }}int main(){ int T,i,j,k; cin>>T; while(T--) { cin>>n>>m; cnt=0; for(i=0;i
                  
                   >ch[i][j]; b[i][j]=INF; if(ch[i][j]=='J') { p=i;q=j; ch[i][j]='.'; } else if(ch[i][j]=='F') { d[cnt][0]=i; d[cnt++][1]=j; } } } qq=0; bfs(); bsf(); if(qq==0) cout<<"IMPOSSIBLE"<