BFS广度优先搜索(10)--fzu2150(基础题）

Fire Game

                                    Time Limit:1000MS    Memory Limit:32768KB    64bit IO Format:%I64d & %I64u

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

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

Sample Output

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

            这道题就是说任选两个是‘#’的格子（‘#‘表示此格子是能够燃烧的草，‘.’ 表示是不能燃烧的石头），同时点燃，一个格子的火一秒钟能够将它上下左右的是草的格子都燃烧掉，问最少经过多少秒将所有是草的格子燃烧完，若不能燃烧完，输出-1。

        解题思路很简单，就是分别枚举两个不相同的格子选中点燃（若只有一个或者两个格子，则直接输出0），枚举完所有情况后再比较输出所用最少的时间。代码如下：

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
char map[12][12];
int vis[12][12];         
int d[4][2]={0,1,1,0,0,-1,-1,0};  //上下左右四个方向
int n,m;
struct node{
	int x,y;
	int step;
};
int fire[12][12];      //已经燃起火的格子为1
node f[100];
int cont;        //是草的格子的数量
int Bfs(node s,node s1){     
	queue<node>q;
	node e;
	int i;
	memset(fire,0,sizeof(fire));
	fire[s.x][s.y]=1;
	fire[s1.x][s1.y]=1;
	q.push(s);
	q.push(s1);
	int num=2;
	while(!q.empty())
	{
		s=q.front();
		q.pop();
		for(i=0;i<4;i++){
			int xx=s.x+d[i][0];
			int yy=s.y+d[i][1];
			if(xx<0||yy<0||xx>=n||yy>=m)
				continue;
			if(map[xx][yy]=='.')continue;
			if(fire[xx][yy])continue;
			fire[xx][yy]=1;
			num++;
			e.x=xx;
			e.y=yy;
			e.step=s.step+1;
			q.push(e);
		}
		if(q.empty()&&num==cont)return s.step;     //所有为草的格子都被燃烧完了
	}
	return -1;       //有为草的格子没有燃烧完
}
int main()
{
	int T,i,j;
	int cnt=0;
	scanf("%d",&T);
	while(T--){
		scanf("%d %d",&n,&m);
		memset(f,0,sizeof(f));
		for(i=0;i<n;i++){
			scanf("%s",map[i]);
		}
		cont=0;
		for(i=0;i<n;i++){                //任选两个是草的格子为起点开始Bfs
			for(j=0;j<m;j++){
				if(map[i][j]=='#'){
					node a;
					a.x=i;
					a.y=j;
					a.step=0;
					f[cont++]=a;
				}
			}
		}
		if(cont==1||cont==2){        //有草的格子只有一个或者两个
			printf("Case %d: 0\n",++cnt);
			continue;
		}
		int ans=105;
		for(i=0;i<cont;i++){               //找出所用时间最小的
			for(j=i+1;j<cont;j++){
				int ans1=Bfs(f[i],f[j]);
				if(ans1==-1)
				{
					if(ans==105)ans=-1;
				}
				else{
					if(ans==-1)ans=ans1;
					else ans=(ans<ans1)?ans:ans1;
				}
			}
		}
		printf("Case %d: %d\n",++cnt,ans);
	}
	return 0;
}