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
这个题一开始想麻烦了，一开始想的是先判断连通块的个数，再在每一个连通块里寻找最长链，于是去想了很久的最长链求
法，然后发现如果是一坨都是草的话就会出问题，看了一眼标程，竟是最朴素的暴力算法，然后去查了一下原来1s可以运行
的次数是5*1e8次，是我孤陋寡闻了一直以为是千万级。
像这种有关最短路问题的搜索一般都是用bfs去做。
程序为了避免卡常还是做了一点操作的qwq
1 #include <cstdio> 2 #include <cmath> 3 #include <cstring> 4 #include <cstdlib> 5 #include <queue> 6 #include <stack> 7 #include <vector> 8 #include <iostream> 9 #include "algorithm" 10 using namespace std; 11 const int MAX=15; 12 int t,cas,n,m,ans; 13 char s[MAX][MAX]; 14 bool vis[MAX][MAX]; 15 int dx[]={1,0,-1,0}; 16 int dy[]={0,1,0,-1}; 17 struct Poi{ 18 int x,y,ste; 19 }; 20 queue <Poi> q; 21 inline int mx(int x,int y){return x>y?x:y;} 22 inline int mn(int x,int y){return x<y?x:y;} 23 void get_pos(int nu,int *x,int *y){ 24 *x=nu/m+1;if (nu%m==0) (*x)--; 25 *y=nu%m;if (nu%m==0) (*y)=m; 26 } 27 void bfs(int cx,int cy,int vx,int vy){ 28 int i,j,cc,vv,an=0;Poi a,b; 29 while (!q.empty()) q.pop(); 30 memset(vis,false,sizeof(vis)); 31 a.x=cx;a.y=cy,a.ste=0; 32 b.x=vx;b.y=vy,b.ste=0; 33 q.push(a);q.push(b); 34 vis[cx][cy]=vis[vx][vy]=true; 35 while (!q.empty()){ 36 a=q.front();q.pop();an=mx(an,a.ste); 37 for (i=0;i<4;i++){ 38 cc=a.x+dx[i]; 39 vv=a.y+dy[i]; 40 if (s[cc][vv]=='#' && !vis[cc][vv]){ 41 b.x=cc,b.y=vv,b.ste=a.ste+1; 42 vis[cc][vv]=true; 43 q.push(b); 44 } 45 } 46 } 47 for (i=1;i<=n;i++) 48 for (j=1;j<=m;j++) 49 if (s[i][j]=='#' && !vis[i][j]) 50 return; 51 ans=mn(ans,an); 52 } 53 int main(){ 54 freopen ("fire.in","r",stdin); 55 freopen ("fire.out","w",stdout); 56 int i,j,cx,cy,vx,vy; 57 scanf("%d",&t); 58 while (t--){ 59 scanf("%d%d ",&n,&m); 60 ans=1e8; 61 for (i=1;i<=n;i++) scanf("%s",s[i]+1); 62 for (i=1;i<=n*m;i++){ 63 get_pos(i,&cx,&cy); 64 if (s[cx][cy]=='#') 65 for (j=i+1;j<=n*m;j++){ 66 get_pos(j,&vx,&vy); 67 if (s[vx][vy]=='#') 68 bfs(cx,cy,vx,vy); 69 } 70 } 71 printf("Case %d: ",++cas); 72 if (ans==1e8) printf("-1 "); 73 else printf("%d ",ans); 74 } 75 return 0; 76 }