I - Fire Game FZU 2150 双起点的BFS

I - Fire Game
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit Status Practice FZU 2150

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

###

..#

#.#

//题意是连通分量小于2的时候，任选两点（可以重合）开始，求最少的时间访问到所有的#位置，每步耗时1个单位，

//就是两个起点的同时搜索，程序里实际上还是交替进行搜索的，以此实现同时的搜索

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<vector>
#include<map>
#include<stack>
using namespace std;
const int maxn=15;
const int inf=200000;
#define lson rt<<1,l,m
#define rson rt<<1|1,m+1,r
#define For(i,n) for(int i=0;i<(n);i++)
template<class T>inline T read(T&x)
{
    char c;
    while((c=getchar())<=32);
    bool ok=false;
    if(c=='-')ok=true,c=getchar();
    for(x=0; c>32; c=getchar())
        x=x*10+c-'0';
    if(ok)x=-x;
    return x;
}
template<class T> inline void read_(T&x,T&y)
{
    read(x);
    read(y);
}
template<class T> inline void write(T x)
{
    if(x<0)putchar('-'),x=-x;
    if(x<10)putchar(x+'0');
    else write(x/10),putchar(x%10+'0');
}
template<class T>inline void writeln(T x)
{
    write(x);
    putchar('\n');
}
//  -------IO template------
bool vis[maxn][maxn];
char G[maxn][maxn];
int n,m;
int dx[]= {1,-1,0,0};
int dy[]= {0,0,1,-1};
void dfs(int x,int y)
{
    vis[x][y]=true;
    for(int i=0; i<4; i++)
    {
        int xx=dx[i]+x;
        int yy=dy[i]+y;
        if(xx>=0&&xx<n&&yy>=0&&yy<m&&G[xx][yy]=='#'&&!vis[xx][yy])
        {
            dfs(xx,yy);
        }
    }
}
struct node
{
    int x,y;
    node(int a,int b)
    {
        x=a;
        y=b;
    }
};
int d[maxn][maxn];
int rec=1;
void bfs(int x,int y,int x1,int y1)
{
    queue<node> q;
    q.push(node(x,y));
    q.push(node(x1,y1));
    vis[x][y]=true;
    vis[x1][y1]=true;
    d[x][y]=0;
    d[x1][y1]=0;
    while(!q.empty())
    {
        node s=q.front();
        q.pop();
        for(int i=0; i<4; i++)
        {
            int xx=dx[i]+s.x;
            int yy=dy[i]+s.y;
            if(xx>=0&&xx<n&&yy>=0&&yy<m&&!vis[xx][yy]&&G[xx][yy]=='#')
            {
                vis[xx][yy]=true;
                q.push(node(xx,yy));
                d[xx][yy]=d[s.x][s.y]+1;
                rec=max(rec,d[xx][yy]);
            }
        }
    }
}

bool ok()
{
    for(int i=0; i<n; i++)
        for(int j=0; j<m; j++)
            if(G[i][j]=='#'&&!vis[i][j])return false;
    return true;
}

bool mmvis[maxn][maxn][maxn][maxn];//双起点的标记

int main()
{
    int i,j,k,t;
    int T;
#ifndef ONLINE_JUDGE
    freopen("test.txt","r",stdin);
    freopen("Ab.txt","w",stdout);
#endif // ONLINE_JUDGE
    read(T);
    int cas=1;
    while(T--)
    {
        memset(G,0,sizeof(G));
        read_(n,m);
        for(i=0; i<n; i++)
            for(j=0; j<m; j++)
                cin>>G[i][j];


        memset(vis,0,sizeof(vis));
        int ans=0;
        For(i,n)For(j,m)if(G[i][j]=='#'&&!vis[i][j])
        {
            dfs(i,j);
            ans++;
        }
        //writeln(ans);
        if(ans>2)ans=-1;
        else
        {
            ans=inf;
            memset(mmvis,false,sizeof(mmvis));
            For(i,n)For(j,m)if(G[i][j]=='#')
            {
                For(u,n)For(v,m)if(G[u][v]=='#'&&!mmvis[i][j][u][v])
                {
                    mmvis[i][j][u][v]=mmvis[u][v][i][j]=true;
                    memset(vis,0,sizeof(vis));
                    memset(d,0,sizeof(d));
                    rec=0;
                    bfs(i,j,u,v);
                    if(ok()&&rec<ans)ans=rec;
                }
            }
        }
        printf("Case %d: %d\n",cas++,ans);
    }
    return 0;
}