FZU 2150 Fire Game 【两点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】：两个孩子在n*m的平地上放火玩，#表示草，两个熊孩子分别选一个#格子点火，火可以向上向下向左向右在有草的格子蔓延，点火的地方时间为0，蔓延至下一格的时间依次加一。求烧完所有的草需要的最少时间。如不能烧完输出-1。
依次枚举两个#格子，BFS出需要的时间，取最小的一个。
【代码】：

#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
#define debug() puts("++++")
#define gcd(a,b) __gcd(a,b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a,b,sizeof(a))
#define sz size()
#define be begin()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
#define all 1,n,1
#define rep(i,x,n) for(int i=(x); i<(n); i++)
#define in freopen("in.in","r",stdin)
#define out freopen("out.out","w",stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e18;
const int maxn = 1e3 + 20;
const int maxm = 1e6 + 10;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int dx[] = {-1,1,0,0,1,1,-1,-1};
const int dy[] = {0,0,1,-1,1,-1,1,-1};
int dir[4][2] = {{0,1},{0,-1},{-1,0},{1,0}};
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int n,m,f,e,l;
char a[20][20];
int v[150][150];
struct node
{
    int x,y,step;
}st,ed,d[150];
bool ok(int x,int y)
{
    return x>=0 && y>=0 && x<n && y<m && !v[x][y] && a[x][y]=='#';
}

int bfs(node a,node b) //两点BFS
{
    ms(v,0);
    queue<node>q;
    v[a.x][a.y]=1;
    v[b.x][b.y]=1;
    q.push(a);
    q.push(b);
    f=0;
    while(!q.empty())
    {
        st = q.front();
        q.pop();
        for(int i=0; i<4; i++)
        {
            ed.x = st.x + dir[i][0];
            ed.y = st.y + dir[i][1];
            if(ok(ed.x,ed.y))
            {
                v[ed.x][ed.y] = 1;
                ed.step = st.step + 1;
                f = max(f,ed.step); //同时烧。选时间长的，因为先烧完也要的那个后烧完的。
                q.push(ed);
            }
        }
    }
    return f;
}
bool check()
{
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            if(a[i][j]=='#' && !v[i][j]) //在草坪但未访问——>还有没烧到的草坪
                return false; //则返回假
        }
    }
    return true;
}
int main()
{
    int t,cas=1;
    scanf("%d",&t);
    while(t--)
    {
        int k = 0;
        scanf("%d%d",&n,&m);
        getchar();
        for(int i=0;i<n;i++)
            scanf("%s",a[i]);
        rep(i,0,n)
        {
            rep(j,0,m)
            {
                if(a[i][j] == '#')
                {
                   d[k].x = i;
                   d[k].y = j;
                   d[k].step = 0;
                   k++; //k个有草坪的地方
                }
            }
        }
       int sum = INF;
       for(int i=0;i<k;i++) //枚举两个草坪组合
       {
           for(int j=0;j<k;j++)
           {
               int ans = bfs(d[i],d[j]); //ans是不同组合所花时间
               if(ans < sum && check()) //后面那个函数用来判断草坪都烧完了，选出最少时间烧完的组合
               {
                   sum = ans;
               }
           }
       }
      printf("Case %d: ",cas++);
      if(sum==INF) puts("-1");
      else printf("%d\n",sum);
    }
}
/*
4
3 3
.#.
###
.#.

3 3
.#.
#.#
.#.

3 3
...
#.#
...

3 3
###
..#
#.#
*/