# FZU - 2150 Fire Game（双路bfs）

Fat brother and Maze are playing a kind of special (hentai) game on an N*M board (N rows, M columns). At the beginning, each grid of this board is consisting of grass or just empty and then they start to fire all the grass. Firstly they choose two grids which are consisting of grass and set fire. As we all know, the fire can spread among the grass. If the grid (x, y) is firing at time t, the grid which is adjacent to this grid will fire at time t+1 which refers to the grid (x+1, y), (x-1, y), (x, y+1), (x, y-1). This process ends when no new grid get fire. If then all the grid which are consisting of grass is get fired, Fat brother and Maze will stand in the middle of the grid and playing a MORE special (hentai) game. (Maybe it’s the OOXX game which decrypted in the last problem, who knows.)

You can assume that the grass in the board would never burn out and the empty grid would never get fire.

Note that the two grids they choose can be the same.

Input

The first line of the date is an integer T, which is the number of the text cases.

Then T cases follow, each case contains two integers N and M indicate the size of the board. Then goes N line, each line with M character shows the board. “#” Indicates the grass. You can assume that there is at least one grid which is consisting of grass in the board.

1 <= T <=100, 1 <= n <=10, 1 <= m <=10

Output

For each case, output the case number first, if they can play the MORE special (hentai) game (fire all the grass), output the minimal time they need to wait after they set fire, otherwise just output -1. See the sample input and output for more details.

Sample Input

```4
3 3
.#.
###
.#.
3 3
.#.
#.#
.#.
3 3
...
#.#
...
3 3
###
..#
#.#```

Sample Output

```Case 1: 1
Case 2: -1
Case 3: 0
Case 4: 2```

#的地方是草，开始时最多点燃两个草块，每个着火的草块在1秒后会蔓延到与其相邻的四个草块上，问最多需要等多少秒可以烧完所有草，不可以的话输出-1

``````#include
#include
#include
#include
#include
#include
using namespace std;
const int inf=0x3f3f3f3f;
int dir= {{1,0},{-1,0},{0,1},{0,-1}};
int mp;
int be;///每次bfs时mp的副本
int n,m,tot,cnt,ans,minn;

struct node
{
int a,b;
node(int a,int b):a(a),b(b){}///构造函数
};

vectorvec;///存草垛的位置

void bfs(node x,node y)///两点同时入队
{
queueq;
q.push(x);
q.push(y);
cnt+=2;
int step=0;
be[x.a][x.b]=0;
be[y.a][y.b]=0;
while(q.size())
{
step++;///新的一层
if(step>=ans)///剪枝
return;
int cn=q.size();
while(cn--)///将此时队里的所有点，即同一层的点全部pop出来，算是一步
{
node tmp=q.front();
q.pop();
for(int i=0; i<4; i++)
{
int fx=tmp.a+dir[i];
int fy=tmp.b+dir[i];
if(fx>=1&&fx<=n&&fy>=1&&fy<=m&&be[fx][fy])
{
be[fx][fy]=0;
q.push(node(fx,fy));
cnt++;///有cnt个草垛被烧过了
if(cnt==tot)
{
ans=step;
return;
}
}
}
}
}
}
int main()
{
int t;
scanf("%d",&t);
for(int k=1; k<=t; k++)
{
vec.clear();
scanf("%d%d",&n,&m);
minn=inf;
tot=0;
char c;
for(int i=1; i<=n; i++)
{
getchar();
for(int j=1; j<=m; j++)
{
c=getchar();
if(c=='.')
mp[i][j]=0;
else if(c=='#')
{
mp[i][j]=1;
tot++;
vec.push_back(node(i,j));
}
}
}
//        cout<<<'\n';>2)
{
for(int i=0; i<<'\n';>``````
```   ``` ### CSDN

5

) 