题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=654

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

PS:

http://blog.csdn.net/acdreamers/article/details/8654005

这图要用邻接表。 否则会爆内存。

50*50/2！ 开1500的二维就OK拉。

代码例如以下：

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<iostream>

using namespace std;

#define MAXN 1547

int LN, RN;//L,R数目

int g[MAXN][MAXN], linker[MAXN];

bool used[MAXN];

char ma[57][57];

int r[MAXN][MAXN], c[MAXN][MAXN];

int head[MAXN];

struct node

{

    int x;

    int next;

} edge[MAXN<<1];

int cnt = 1;

void addEdge(int x, int y)

{

    edge[cnt].x = y;

    edge[cnt].next = head[x];

    head[x] = cnt++;

}

int dfs(int L)//从左边開始找增广路径

{

    int R;

    //for(R = 1; R <= RN; R++)

    for(int i = head[L]; ~i; i = edge[i].next)

    {

        R = edge[i].x;

        if(!used[R])

        {

            //找增广路。反向

            used[R]=true;

            if(linker[R] == -1 || dfs(linker[R]))

            {

                linker[R] = L;

                return 1;

            }

        }

    }

    return 0;

}

int hungary()

{

    int res = 0 ;

    int L;

    memset(linker,-1,sizeof(linker));

    for( L = 1; L <= LN; L++)

    {

        memset(used,0,sizeof(used));

        if(dfs(L) != 0)

            res++;

    }

    return res;

}

int check(int x, int y)

{

    if(ma[x][y]=='o' || ma[x][y]=='*')

    {

        return 1;

    }

    return 0;

}

void init()

{

    cnt = 0;

    memset(head,-1,sizeof(head));

    memset(g,0,sizeof(g));

    memset(c,0,sizeof(c));

    memset(r,0,sizeof(r));

}

int main()

{

    int t;

    int n, m;

    int k, L, R;

    int cas = 0;

    scanf("%d",&t);

    while(t--)

    {

        init();

        scanf("%d%d",&n,&m);

        for(int i = 1; i <= n; i++)

        {

            scanf("%s",ma[i]+1);

        }

        int cnt1 = 1, cnt2 = 1;

        for(int i = 1; i <= n; i++)

        {

            for(int j = 1; j <= m; j++)

            {

                if(check(i, j))

                {

                    if(r[i][j] == 0)//行

                    {

                        for(int k = j; k <= m; k++)

                        {

                            if(check(i, k))

                            {

                                r[i][k] = cnt1;

                            }

                            else

                            {

                                break;

                            }

                        }

                        cnt1++;

                    }

                    if(c[i][j] == 0)//列

                    {

                        for(int k = i; k <= n; k++)

                        {

                            if(check(k, j))

                            {

                                c[k][j] = cnt2;

                            }

                            else

                            {

                                break;

                            }

                        }

                        cnt2++;

                    }

                }

            }

        }

        for(int i = 1; i <= n; i++)

        {

            for(int j = 1; j <= m; j++)

            {

                if(r[i][j] && c[i][j] && ma[i][j]=='o')

                {

                    //g[r[i][j]][c[i][j]] = 1;

                    //g[c[i][j]][r[i][j]] = 1;

                    addEdge(r[i][j], c[i][j]);

                }

            }

        }

        LN = cnt1;

        RN = cnt2;

        int ans = hungary();

        printf("Case :%d\n",++cas);

        printf("%d\n",ans);

    }

    return 0 ;

}

/*

2

4 4

o***

*###

oo#o

***o

4 4

#ooo

o#oo

oo#o

***#

*/