POJ 3020Antenna Placement(二分图最小顶点覆盖)

Antenna Placement

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5649 Accepted: 2828

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**oo
oo**oo*
ooo*o*o
o**o**o
ooooooo
o******
*ooo*o*
oo*oo**
*****oo
10 1
*
*
*
o
*
*
*
*
*
*

Sample Output

17
5

 

/***************************************************************** 题目大意：一个矩形中，有N个城市’*’，现在这n* 个城市都要覆盖无线，若放置一个基站，那么它至* 多可以覆盖相邻的两个城市。* 问至少放置多少个基站才能使得所有的城市都覆盖无线？* 分析：就是一个最小顶点覆盖的问题；* 解答：可以通过转化为求最大匹配来解决；* 利用公式 最小顶点覆盖=结点数量-最大匹配/2；** 建图方法：要明确的是，输入的一堆“圈圈星星”可以* 看做是一张大地图，地图上有所有城市的坐标，但是这里有一个误区：* 不能简单地把城市的两个x、y坐标作为准备构造的二分图的两个顶点集。* 城市才是要构造的二分图的顶点！* 此处分析转自：http://blog.csdn.net/lyy289065406/article/details/6647040* 构造方法如下：* 例如输入：* *oo* **** O*o* 时，可以抽象为一个数字地图：* 100* 234* 050*数字就是根据输入的城市次序作为该城市的编号，0代表该位置没有城市。*然后根据题目的“范围”规则，从第一个城市开始，以自身作为中心城市，向四个方向的城市进行连线（覆盖）* 因此就能够得到边集：edge(1,2),edge(2,1),edge(3,2),edge(2,3),edge(3,4),edge(4,3),edge(3,5),edge(5,3);* 下来套模板做就OK了；*   author: crazy_石头*   备注：邻接表版本*   Problem: POJ3020--Antenna Placement***************************************************************************/#include <iostream>#include <cstdlib>#include <cstdio>#include <cstring>#include <algorithm>#define A system("pause")using namespace std;const int maxn=400+5;char map[maxn][maxn];int city[maxn][maxn];int Linker[maxn],head[maxn];bool used[maxn];int dx[]={-1,0,1,0};int dy[]={0,-1,0,1};int n;//城市数量；int uN,vN;//左右点集；int res;//最大匹配；int cnt,h,w;struct Node{    int v,next;}edge[maxn<<2];bool OK(int x,int y){    if(x<0||x>=h||y<0||y>=w||map[x][y]!='*')        return false;    return true;}void addedge(int u,int v){     edge[cnt].v=v;     edge[cnt].next=head[u];     head[u]=cnt++;}int DFS(int u){     for(int i=head[u];~i;i=edge[i].next)     {         int v=edge[i].v;         if(!used[v])         {             used[v]=1;             if(Linker[v]==-1||DFS(Linker[v]))             {                 Linker[v]=u;                return 1;             }         }     }     return 0;}int Hungary(){     int res=0;     memset(Linker,-1,sizeof(Linker));     for(int i=0;i<uN;i++)     {         memset(used,0,sizeof(used));        if(DFS(i))             res++;     }     return res;}int main(){     int test;     scanf("%d",&test);     while(test--)     {         cnt=0;         memset(head,-1,sizeof(head));         n=0;         scanf("%d%d%*c",&h,&w);         for(int i=0;i<h;i++)         scanf("%s",map[i]);        //A;         for(int i=0;i<h;i++)         {             for(int j=0;j<w;j++)             {                 if(map[i][j]=='*')                 {                     city[i][j]=n++;                    //printf("%d\n",city[i][j]);                    //cout<<12345<<endl;                    //A;                 }             }         }         //A;         for(int i=0;i<h;i++)         {             for(int j=0;j<w;j++)             {                if(map[i][j]!='*')                    continue;                 for(int k=0;k<4;k++)                 {                    int xx=i+dx[k];                    int yy=j+dy[k];                    if(!OK(xx,yy))                        continue;                     addedge(city[i][j],city[xx][yy]);                 }             }         }         uN=n;         int res=uN-Hungary()/2;         printf("%d\n",res);     }     //A;     return 0;}/******************************************* 这个题目中又出现了两处小错误。* 1.手误，把+敲成了=。。* 2.把++n和n++用错了，幸亏改好了1Y* 以后一定要细心啊******************************************/