Place the Robots

Robert is a famous engineer. One day he was given a task by his boss. The background of the task was the following:

Given a map consisting of square blocks. There were three kinds of blocks: Wall, Grass, and Empty. His boss wanted to place as many robots as possible in the map. Each robot held a laser weapon which could shoot to four directions (north, east, south, west) simultaneously. A robot had to stay at the block where it was initially placed all the time and to keep firing all the time. The laser beams certainly could pass the grid of Grass, but could not pass the grid of Wall. A robot could only be placed in an Empty block. Surely the boss would not want to see one robot hurting another. In other words, two robots must not be placed in one line (horizontally or vertically) unless there is a Wall between them.

Now that you are such a smart programmer and one of Robert’s best friends, He is asking you to help him solving this problem. That is, given the description of a map, compute the maximum number of robots that can be placed in the map.

Input

The first line contains an integer T (<= 11) which is the number of test cases.

For each test case, the first line contains two integers m and n (1<= m, n <=50) which are the row and column sizes of the map. Then m lines follow, each contains n characters of ‘#’, ‘*’, or ‘o’ which represent Wall, Grass, and Empty, respectively.

Output

For each test case, first output the case number in one line, in the format: “Case :id” where id is the test case number, counting from 1. In the second line just output the maximum number of robots that can be placed in that map.

Sample Input

2
4 4
o***
*###
oo#o
***o
4 4
#ooo
o#oo
oo#o
***#

Sample Output

Case :1
3
Case :2
5

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分析
我们将每一行，每一列被墙隔开，且包含空地的连续区域称作“块”。显然，在一个块之中，最多只能放一个机器人。我们把这些块编上号。
同样，把竖直方向的块也编上号。
把每个横向块看作X部的点，竖向块看作Y部的点，若两个块有公共的空地，则在它们之间连边。
由于每条边表示一个空地，有冲突的空地之间必有公共顶点，所以问题转化为二部图的最大匹配问题。
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程序：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
 
int n,m;
char ch[55][55];
int x[2509],y[2509],xs[55][55],ys[55][55],xn,yn,s[2509],nt[2509],e[2509],visit[2509];
 
bool path(int k)
{
    for (int i=s[k];~i;i=nt[i])
    {
        int ee=e[i];
        if (!visit[ee])
        {
            visit[ee]=1;
            if (y[ee]==-1||path(y[ee]))
            {
                y[ee]=k;
                x[k]=ee;
                return 1;
            }
        }
    }
    return 0;
}
 
void work()
{
    int ans=0;
    memset(x,-1,sizeof x);
    memset(y,-1,sizeof y);
    for (int i=1;i<=xn;i++)
    {
        if (x[i]==-1)
        {
            memset(visit,0,sizeof visit);
            if (path(i)) ans++;
        }
    }
    printf("%d
",ans);
}
 
int main()
{
    int t,cas=0;
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d%d",&n,&m);
        int flag=0,cnt=0;
        for (int i=0;i<n;i++)
            scanf("%s",ch[i]);
        memset(xs,0,sizeof xs);
        memset(ys,0,sizeof ys);
        for (int i=0;i<n;i++)
        {
            flag=0;
            for (int j=0;j<m;j++)
            {
                if (ch[i][j]=='o')
                {
                    if (!flag) cnt++;
                    xs[i][j]=cnt;
					flag=1;
                } else 
				if (ch[i][j]=='#') flag=0;
            }
        }
        xn=cnt;
		cnt=0;
        for (int i=0;i<m;i++)
        {
            flag=0;
            for (int j=0;j<n;j++)
            {
                if (ch[j][i]=='o')
                {
                    if (!flag) cnt++;
                    ys[j][i]=cnt;
					flag=1;
                } else 
				if (ch[j][i]=='#') flag=0;
            }
        }
        yn=cnt;
		cnt=1;
        memset(s,-1,sizeof s);
        memset(nt,-1,sizeof nt);
        for (int i=0;i<n;i++)
        {
            for (int j=0;j<m;j++)
            {
                if (xs[i][j]&&ys[i][j])
                {
                	nt[cnt]=s[xs[i][j]];
					s[xs[i][j]]=cnt;
					e[cnt++]=ys[i][j];
                }
            }
        }
        printf("Case :%d
",++cas);
        work();
    }
    return 0;
}