Description

After escaping from Figo's chase, Severus falls into a N * M maze designed by Figo.

At first, Severus is located on the grid S. Every second he can only move to the four grids that adjacent to the grid he is

 located on. The moment he move to any side of the maze, he will get rid of Figo.

After T seconds, Figo will reach the maze. Because Figo is the designer of the maze, when Figo arrive, he can reach any 

grid if he want. If Severus can't leave the maze at that moment, there is no doubt that he will be caught by Figo.

Figo is very cunning. In the maze he set not only walls, but also fire! After every second, the fire will spread to the four 

grid that adjacent to it. When a grid is on fire, certainly, Severus can not be on the grid. Can Severus escape from the

 maze?

Input

The first line will be a integer CS, indicating the number of test cases.

In every case, there will be three integer N, M, T.

After that there will be N * M characters, decribe the maze.

The "." is a empty grid, "#" is wall, "F" is the fire, "S" is the initial grid that Severus stands on.

10 <= n , m  <= 100      10 <= T <=10000

Output

There is only one line, if Severus can get out, output the mininum time he need to escape from the maze. If he can't,

 output "Poor Severus must be caught by strong Figo!!!"

Sample Input

2

4 4 4

####

#SF#

#..#

#..#

3 3 4

###

#S.

#.F

 

Sample Output

3

Poor Severus must be caught by strong Figo!!!

 

Source

2014 Winter Holiday Contest 5

Author

TwIStOy

思路：双重bfs  只需要两个vis分别标记并记录路径  q1 q2 分别储存点即可

代码：

1 #include
      
         2 #include
       
          3 #include
        
           4 #include
         
            5 using namespace std;  6   7 struct point  8 {  9     int x; 10     int y; 11 }poin[10]; 12  13 queue
          
           q1,q2; 14  15 int xx[4]={
    0,0,1,-1}; 16 int yy[4]={
    1,-1,0,0}; 17  18 int n,m; 19  20 const int maxn=105; 21  22 int vis1[maxn][maxn]; 23 int vis2[maxn][maxn]; 24 char a[maxn][maxn]; 25  26 int bfs() 27 { 28     while(!q1.empty()) 29     { 30         while(!q2.empty()&&vis2[q2.front().x][q2.front().y]!=vis1[q1.front().x][q1.front().y]) 31         { 32             int tmx=q2.front().x; 33             int tmy=q2.front().y; 34             q2.pop(); 35             for(int j=0;j<4;j++) 36             { 37                 poin[3].x=tmx+xx[j]; 38                 poin[3].y=tmy+yy[j]; 39                 if(!vis2[poin[3].x][poin[3].y]&&poin[3].x>=0&&poin[3].x
           
            =0&&poin[3].y
            
             =n||poin[4].y<0||poin[4].y>=m) 56 return vis1[tmpx][tmpy]+1; 57 58 59 vis1[poin[4].x][poin[4].y]=vis1[tmpx][tmpy]+1; 60 q1.push(poin[4]); 61 } 62 } 63 } 64 return 0xfffffff; 65 } 66 67 int main() 68 { 69 int t; 70 int time; 71 ios::sync_with_stdio(false); 72 //freopen("aa.txt","r",stdin); 73 cin>>t; 74 while(t--) 75 { 76 while(!q1.empty()) 77 q1.pop(); 78 while(!q2.empty()) 79 q2.pop(); 80 81 cin>>n>>m>>time; 82 for(int i=0;i
             
              >a[i][j]; 85 if(a[i][j]=='S') 86 { 87 poin[0].x=i; 88 poin[0].y=j; 89 vis1[i][j]=1; 90 q1.push(poin[0]); 91 } 92 else if(a[i][j]=='F') 93 { 94 poin[1].x=i; 95 poin[1].y=j; 96 vis2[i][j]=1; 97 q2.push(poin[1]); 98 } 99 }100 }101 int bx=bfs()-1;102 if(bx<=time)103 cout<
              
               <