POJ 3020 Antenna Placement

Description

The Global Aerial Research Centre has been allotted the task of building the fifth generation of mobile phone nets in Sweden. The most striking reason why they got the job, is their discovery of a new, highly noise resistant, antenna. It is called 4DAir, and comes in four types. Each type can only transmit and receive signals in a direction aligned with a (slightly skewed) latitudinal and longitudinal grid, because of the interacting electromagnetic field of the earth. The four types correspond to antennas operating in the directions north, west, south, and east, respectively. Below is an example picture of places of interest, depicted by twelve small rings, and nine 4DAir antennas depicted by ellipses covering them.

Obviously, it is desirable to use as few antennas as possible, but still provide coverage for each place of interest. We model the problem as follows: Let A be a rectangular matrix describing the surface of Sweden, where an entry of A either is a point of interest, which must be covered by at least one antenna, or empty space. Antennas can only be positioned at an entry in A. When an antenna is placed at row r and column c, this entry is considered covered, but also one of the neighbouring entries (c+1,r),(c,r+1),(c-1,r), or (c,r-1), is covered depending on the type chosen for this particular antenna. What is the least number of antennas for which there exists a placement in A such that all points of interest are covered?

Input

On the first row of input is a single positive integer n, specifying the number of scenarios that follow. Each scenario begins with a row containing two positive integers h and w, with 1 <= h <= 40 and 0 < w <= 10. Thereafter is a matrix presented, describing the points of interest in Sweden in the form of h lines, each containing w characters from the set ['*','o']. A '*'-character symbolises a point of interest, whereas a 'o'-character represents open space.

Output

For each scenario, output the minimum number of antennas necessary to cover all '*'-entries in the scenario's matrix, on a row of its own.

Sample Input

2
7 9
ooo**oooo
**oo*ooo*
o*oo**o**
ooooooooo
*******oo
o*o*oo*oo
*******oo
10 1
*
*
*
o
*
*
*
*
*
*

Sample Output

17
5

Source

Svenskt Mästerskap i Programmering/Norgesmesterskapet 2001

/*
   匈牙利算法的模板题，其实我都没想到可以A掉。
   题目大意: 给定一个图，所有的‘ * ’是一个个的基站，而‘ o ’则是基站之间的空地，现在在地图上你可以用圈把基站圈起来，每次只能圈1<=x<=2个，问最 少需要圈多少次能够把所有的基站全都圈进来。

   解题思路:还是模板题，而重要的是将给定的地图或坐标或其他东西，转化为我们想要的二分图。
*/

#include <algorithm>
#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <cstdio>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
using namespace std;
const int INF = 0x3f3f3f3f;
const int M = 560;
 
int dx[] = {0,0,1,-1};
int dy[] = {1,-1,0,0};
char str[45][15];
bool Map[M][M],vis[M];
int dis[M],n,m;

bool f(int s)
{
    for(int j=1;j<=n*m;j++)
    {
        if(Map[s][j] && !vis[j])
        {
            vis[j] = true;
            if(!dis[j] || f(dis[j]))
            {
                dis[j] = s;
                return true;
            }
        }
    }
    return false;
}

int main()
{
    int t,sum,ans;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d",&n,&m);sum = ans = 0;
        memset(Map,false,sizeof(Map));
        memset(dis,0,sizeof(dis));
        for(int i=0;i<n;i++)
            scanf("%s",str[i]);
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(str[i][j] == '*')
                {
                    sum++;
                    for(int k=0;k<4;k++)
                    {
                        int lx = i + dx[k];
                        int ly = j + dy[k];
                        if(lx>=0 && ly>=0 && lx<n && ly<m && str[lx][ly]=='*')
                        {
                            Map[i*m+j+1][lx*m+ly+1] = true;
                            Map[lx*m+ly+1][i*m+j+1] = true;
                        }
                    }
                }
            }
        }
        for(int i=1;i<=n*m;i++)
        {
            memset(vis,false,sizeof(vis));
            if(f(i))
                ans++;
        }
        printf("%d\n",(sum*2-ans)/2);//值得注意的是这里，匈牙利算法求出的是所有点中每两个点为一个集合的最少集合数，sum是所有基站的个数，将基站*2之后减去集合的个数/2，便是我们的答案
    }
    return 0;
}