FZU Problem 2150 Fire Game （BFS，java）

Problem 2150 Fire Game

Accept: 2070    Submit: 7235
Time Limit: 1000 mSec    Memory Limit : 32768 KB

Problem Description

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

43 3.#.###.#.3 3.#.#.#.#.3 3...#.#...3 3###..##.#

Sample Output

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

题目链接：http://acm.fzu.edu.cn/problem.php?pid=2150
解题思路，典型的BFS，枚举两个点，当两个点都满足“点火”条件时，两个点都压入队列（如果两个点位置重合，压入一个点即可）。每个点结构的成员变量有横坐标x、纵坐标y和步数cnt。

代码如下：

import java.util.*;

public class Main{
	public static int T,n,m;
	public static int[] dx = {0,-1,0,1};      //改变上下左右四个方向的数组
	public static int[] dy = {-1,0,1,0};
	public static boolean flog= false;                    //flog用来标记是否能“完全燃烧”矩阵里的草
	public static char[] ch[] = new char[12][12];         //矩阵模型
	public static boolean[] vis[] = new boolean[12][12];  //标记是否访问过当前位置
	static class node{            //每个点的结构体
		int x;
		int y;
		int cnt;
		public node(int x,int y,int cnt){
			this.x = x;
			this.y = y;
			this.cnt = cnt;
		}
	}
	public static int BFS(int x1,int y1,int x2,int y2,int number){
		Queue <node> q = new LinkedList<node>();      //注意队列的初始化一定要写在BFS中
		q.add(new node(x1,y1,0));                     //把初始位置的点压入队列
		if(x1!=x2 || y1!=y2){
			q.add(new node(x2,y2,0));
		}
		int curans = 0;
		while(!q.isEmpty()){
			node nd = q.poll();                       //取出队首的点并把点从队列中移除
			boolean f = true;                         //用来判断在该点之后是否能继续走下去
			for(int i=0;i<4;i++){
				int xx = nd.x+dx[i];
				int yy = nd.y+dy[i];
				int ct = nd.cnt+1;
				if(xx < 0 || yy<0 || xx>=n ||yy>=m|| ch[xx][yy]=='.' || vis[xx][yy]){
					continue;                        //判断下一个点是否出界以及是否满足条件
				}
				vis[xx][yy] = true;
				f = false;
				number--;                          //每走完一步，剩余种草的位置个数减1
				if(number==0){                     //燃烧完所有的草，返回最大步数
					flog = true;
					curans = Math.max(curans,ct);
					return curans;
				}
				node dd = new node(xx,yy,ct);
				q.add(dd);                //把新的点压入队列
			}
			if(f){                        //不能继续走下去，纪录当前点的步数
				curans = nd.cnt;
			}
		}
		return Integer.MAX_VALUE;
	}
	
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		T = sc.nextInt();

		for(int ca = 1;ca <= T;ca ++){
			n = sc.nextInt();
			m = sc.nextInt();
			int num = 0;
			String str = new String();
			for(int i=0;i<n;i++){
				str = sc.next();
				for(int j=0;j<str.length();j++){
					ch[i][j] = str.charAt(j);
					if(ch[i][j]=='#'){
						num++;
					}
				}
			}
			if(num<=2){                   //种草的位置不多于2格，0步即可燃烧完
				System.out.println("Case "+ca+": 0");
				continue;
			}
			int ans = Integer.MAX_VALUE;    //初始化ans的值为最大值
			for(int i=0;i<n;i++){
				for(int j=0;j<m;j++){
					if(ch[i][j]=='.'){
						continue;
					}
					for (int h = i;h<n;h++){
						for(int k=0;k<m;k++){
							if(ch[h][k]=='.'){
								continue;
							}
							flog = false;                  //初始化flog的值
							for(int i1=0;i1<12;i1++){      //初始化标记数组vis
								for(int i2=0;i2<12;i2++){
									vis[i1][i2] = false;
								}
							}
							vis[i][j] = true;
							vis[h][k] = true;
						    int cur = BFS(i,j,h,k,num-2);
							if(flog){
								ans = Math.min(ans,cur);   //成功燃烧完一次，更新ans为最小值
							}
						}
					}
				}
			}
			if(ans==Integer.MAX_VALUE){
				System.out.println("Case "+ca+": -1");
			}else{
				System.out.println("Case "+ca+": "+ans);
			}
		}
	}
}