POJ3020 Antenna Placement(二分图最小路径覆盖)

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

27 9ooo**oooo**oo*ooo*o*oo**o**ooooooooo*******ooo*o*oo*oo*******oo10 1***o******

Sample Output

175

思路

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？

这道题跟上一道题(HDU4185)几乎是一样的，我们依然先给城市进行编号，然后用它来建立二分图，因为是要把整个城市都覆盖完

所以这个题目求的是二分图的最小路径覆盖

有一个定理：

无向二分图的最小路径覆盖 = 顶点数 – 最大二分匹配数/2

所以我们最后用顶点数减去最大二分匹配数就行了

代码

#include<cstdio>#include<cstring>#include<string>#include<set>#include<iostream>#include<stack>#include<queue>#include<vector>#include<algorithm>#define mem(a,b) memset(a,b,sizeof(a))#define inf 0x3f3f3f3f#define mod 10000007#define debug() puts("what the fuck!!!")#define ll long longusing namespace std;const int N=1000+20;int e[N][N],vis[N],match[N],n,m,num,temp[N][N];char s[N][N];int dfs(int u){    for(int i=1; i<=num; i++)    {        if(e[u][i]&&!vis[i])        {            vis[i]=1;            if(!match[i]||dfs(match[i]))            {                match[i]=u;                return 1;            }        }    }    return 0;}int query(){    mem(match,0);    int sum=0;    for(int i=1; i<=num; i++)    {        mem(vis,0);        if(dfs(i))sum++;    }    return sum;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        mem(temp,0);        mem(e,0);        num=0;        scanf("%d%d",&n,&m);        for(int i=1; i<=n; i++)        {            scanf("%s",s[i]+1);            for(int j=1; j<=m; j++)                if(s[i][j]=='*')                    temp[i][j]=++num;        }        for(int i=1; i<=n; i++)            for(int j=1; j<=m; j++)            {                if(s[i][j]=='*')                {                    if(i!=1&&s[i-1][j]=='*') e[temp[i][j]][temp[i-1][j]]=1;                    if(j!=1&&s[i][j-1]=='*') e[temp[i][j]][temp[i][j-1]]=1;                    if(i!=n&&s[i+1][j]=='*') e[temp[i][j]][temp[i+1][j]]=1;                    if(j!=m&&s[i][j+1]=='*') e[temp[i][j]][temp[i][j+1]]=1;                }            }        printf("%d\n",num-query()/2);    }    return 0;}