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
題意:題目的意思是，兩個人在玩游戲之前，需要把草堆全部清除掉，他們兩個可以選擇同一個草堆點燃，也可以選擇不同的（但只能點燃一次，此外火勢是可以蔓延的時間為1分鐘），
問是否可以清除（燃燒）所有的草堆，若能的話輸出做小的燃燒時間，不能就請輸出-1，
思路：首先草堆的數量小于等于2，肯定是能燒完的，此時時間輸出為0
其他情況：開始的兩個著火點可以相同或者不同，所以我們對所有的草堆進行兩兩組合，以這兩個草堆為著火點同時進行廣搜
如果燃燒的草堆數量等于草堆數量，則回傳時間；否則回傳-1
注意：代碼實作有很多注意事項，詳見code
代碼：
import java.util.ArrayDeque; import java.util.Arrays; import java.util.Scanner; class Node{ int x; int y; int step; public Node(int x,int y,int step){ this.x=x; this.y=y; this.step=step; } } public class Main{ static int n,m,c,cnt; static final int N=15; static final int INF=2147483647; static char map[][]=new char[N][N]; static boolean vis[][]=new boolean[N][N]; static ArrayDeque<Node> q=new ArrayDeque<Node>(); static Node node[]=new Node[N*N]; static int dx[]={0,0,1,-1}; static int dy[]={1,-1,0,0}; static int bfs(Node t1,Node t2){ //每次廣搜每次初始化、清空 for(int i=0;i<N;i++) Arrays.fill(vis[i], false); while(!q.isEmpty()) q.poll(); q.offer(new Node(t1.x,t1.y,0)); q.offer(new Node(t2.x,t2.y,0)); vis[t1.x][t1.y]=true; vis[t2.x][t2.y]=true; //必須判斷是否為同1個點 if(t1.x==t2.x && t1.y==t2.y) cnt=1; else cnt=2; // System.out.println("t1.x="+t1.x+" t1.y="+t1.y+" cnt="+cnt); while(!q.isEmpty()){ Node t=q.poll(); for(int i=0;i<4;i++){ int xx=t.x+dx[i]; int yy=t.y+dy[i]; if(xx<0||yy<0||xx>=n||yy>=m||vis[xx][yy]||map[xx][yy]!='#') continue; vis[xx][yy]=true; cnt++; if(cnt==c) return t.step+1; q.offer(new Node(xx,yy,t.step+1)); } } return -1; } public static void main(String[] args) { Scanner scan=new Scanner(System.in); int T=scan.nextInt(); for(int k=1;k<=T;k++){ n=scan.nextInt(); m=scan.nextInt(); for(int i=0;i<n;i++) map[i]=scan.next().toCharArray(); c=0; for(int i=0;i<n;i++) for(int j=0;j<m;j++) if(map[i][j]=='#'){ node[c++]=new Node(i,j,0); } if(c<=2) { System.out.println("Case "+k+": 0"); continue; } int ans=INF; for(int i=0;i<c;i++) for(int j=i;j<c;j++){ int res=bfs(node[i],node[j]);//這里錯搞了2次bfs if(res!=-1) ans=Math.min(ans, res); } if(ans==INF) System.out.println("Case "+k+": -1"); else System.out.println("Case "+k+": "+ans); } } }
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