FZU 2150 Fire Game （双起点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

题意：玩一个点火游戏，#的是草坪，可以点燃而且每秒可以向四周传播，可以选择两个起点开始点火，如果能把草烧完，就输出最小花费的时间，如果不能从两个起点开始把草烧完，那么输出-1。

思路：把每一个草堆存成结构体，包括坐标，和第几步。然后把所有草堆存入vector，暴力枚举两个起点，然后在BFS时同时把两个起点加入队列，返回最后一个出队列的点的步数，然后判断是否合法，即每个草堆都被点燃。

代码如下：

#include<cstdio>
#include<iostream>
#include<queue>
#include<vector>
#include<cstring>
#include<cmath>
#define INF 0x3f3f3f3f
using namespace std;
char mp[15][15];
int vis[15][15];
int dir[4][2]={1, 0, -1, 0, 0, 1, 0, -1};
int m, n;
struct node{
	int x, y, step;
};
vector<node> grass;
bool judge(){//判断是否所有草堆都被点燃 
	for(int i = 0; i < n; i++){
		for(int j = 0; j < m; j++){
			if(mp[i][j]=='#' && !vis[i][j]){
				return false;
			}
		}
	}
	return true;
}
int BFS(node n1, node n2){//双起点BFS 
	int d = 0;
	queue<node> q;
	q.push(n1); 
	q.push(n2);
	while(!q.empty()){
		node nd1 = q.front();
		q.pop();
		if(vis[nd1.x][nd1.y])continue;
		vis[nd1.x][nd1.y] = 1;
		d = nd1.step;//取最后一次出对列的节点的步数 
		for(int i = 0; i < 4; i++){
			int xx = nd1.x + dir[i][0];
			int yy = nd1.y + dir[i][1];
			if(xx >= 0 && xx < n && yy >= 0 && yy < m && !vis[xx][yy] && mp[xx][yy] == '#'){
				q.push((node){xx, yy, nd1.step + 1});
			}
		}
	}
	return d;
}
int main(){
	int cases, i, j, ans, t = 0;
	cin>>cases;
	while(cases--){
		cin>>n>>m;
		t++;
		ans = INF;
		grass.clear();
		memset(vis, 0, sizeof(vis));
		for(i = 0; i < n; i++){
			scanf("%s", mp[i]); 
		}
		for(i = 0; i < n; i++){
			for(j = 0; j < m; j++){
				if(mp[i][j] == '#'){
					grass.push_back((node){i, j, 0});//将草堆结构体存入vector 
				}
			}
		}
		for(i = 0; i < grass.size(); i++){
			for(j = i; j < grass.size(); j++){
				memset(vis, 0, sizeof(vis));//记得初始化 
				grass[i].step = 0;//初始化 
				grass[j].step = 0;
				int temp = BFS(grass[i], grass[j]);//枚举每个起点进行BFS 
				if(judge()){//如果草堆都被点燃 
					ans = min(ans, temp);//取最小值 
				}
			}
		}
		if(ans == INF){//如果该值没有被刷新过 
			ans = -1;
		}
		printf("Case %d: %d\n", t, ans);
	}
	return 0;
}