Fire Game

Problem DescriptionFat 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.InputThe 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 <=10OutputFor 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 Input43 3.#.###.#.3 3.#.#.#.#.3 3...#.#...3 3###..##.#Sample OutputCase 1: 1Case 2: -1Case 3: 0Case 4: 2//题意：给出N*M的矩阵，把所有的'#'变成'.',至多同时两个点扩展,问至少需要多少步才能完成。(选取的初始点不算步数)。//关键字：双向广搜.//标程：#include<iostream>#include<queue>#include<cstdio>#include<cstring>using namespace std;struct node{int x, y, step;};const int inf = 1 << 30;int a[12][12], ans, b[12][12], n, m;queue<node> q;int dir[4][2] = {1,0,0,1,-1,0,0,-1};void bfs(int x,int y,int xx,int yy){b[x][y] = 1, b[xx][yy] = 1;int Max = -1, i, j;while(!q.empty()) q.pop();node s, s1;s.x = x, s.y = y, s.step = 0;s1.x = xx, s1.y = yy, s1.step = 0;q.push(s),  q.push(s1);    while(!q.empty()){        node tmp = q.front();if(tmp.step > Max)  Max = tmp.step; q.pop();for(i = 0; i < 4; ++ i){node now;now.x = dir[i][0] + tmp.x;now.y = dir[i][1] + tmp.y;if(b[now.x][now.y] == 0 && now.x >= 0 && now.x < n && now.y >= 0 && now.y < m){                b[now.x][now.y] = 1;now.step = tmp.step + 1;q.push(now);}}} for(i = 0; i < n; ++ i)for(j = 0; j <m; ++ j)if(b[i][j] == 0) return ;if(ans > Max) ans = Max;}void copy(){for(int i = 0; i < n; ++ i)for(int j = 0; j < m; ++ j)b[i][j] = a[i][j];}int main(){// freopen("a.txt","r",stdin);    int t, i, j, ii, jj;scanf("%d",&t);for(int Cas = 1; Cas <= t; ++ Cas){scanf("%d%d",&n,&m);char ch;memset(a,0,sizeof(a));for(i = 0; i < n; ++ i){getchar();for(j = 0; j < m; ++ j){scanf("%c",&ch);if(ch == '.') a[i][j] = 1;else a[i][j] = 0;}}    ans = inf;for(i = 0; i < n; ++ i)for(j = 0; j < m; ++ j)for(ii = 0; ii < n; ++ ii)for(jj = 0; jj < m; ++ jj){if(!a[i][j] && !a[ii][jj]){copy();bfs(i,j,ii,jj);}}if(ans==inf) ans=-1;printf("Case %d: %d\n",Cas,ans);}return 0;}