POJ 3020 Antenna Placement【二分匹配——最小路徑覆蓋】

連接：

http://poj.org/problem?id=3020

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22010#problem/M

Antenna Placement

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5500		Accepted: 2750

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

題意：

        ‘*’表明城市
       'o'表明空地
       要求你用最小的無線電雷達來覆蓋全部的城市
       雷達只能夠創建在城市上
       每個雷達，只能夠覆蓋它本身所在的位置和與它相鄰的一個位置
       【上 || 下 || 左】
       注意：也就是說一個雷達只能夠覆蓋兩個城市

       英文題目太坑了。。。

算法：二分圖的匹配 ———— 最小路徑覆蓋

      最小路徑覆蓋=最小路徑覆蓋＝｜G｜－最大匹配數

      二分圖的最大匹配總結

思路：

      拆點,把每個城市當作是兩個點
      具體分析仍是看這篇博客學去吧，不能比她說的更好了Orz
      http://blog.csdn.net/lyy289065406/article/details/6647040

PS：感受仍是分不怎麼清何時用最小路徑覆蓋,何時用最大匹配
        (╯▽╰)╭先記住直接匹配的就用最大匹配

        要拆點的就用最小路徑覆蓋好了,水菜傷不起

code：

3020

Accepted

808K

32MS

C++

1926B

/*********************************************************************
題意：‘*’表明城市
       'o'表明空地
       要求你用最小的無線電雷達來覆蓋全部的城市
       雷達只能夠創建在城市上
       每個雷達，只能夠覆蓋它本身所在的位置和與它相鄰的一個位置
       【上 || 下 || 左】
       注意：也就是說一個雷達只能夠覆蓋兩個城市

算法：二分圖的匹配 ———— 最小路徑覆蓋
      最小路徑覆蓋=最小路徑覆蓋＝｜G｜－最大匹配數

思路：拆點,把每個城市當作是兩個點
      具體分析仍是看這篇博客學去吧，不能比她說的更好了Orz
      http://blog.csdn.net/lyy289065406/article/details/6647040
PS：感受仍是分不怎麼清何時用最小路徑覆蓋,何時用最大匹配
    ╮(╯▽╰)╭先記住直接匹配的就用最大匹配
    要拆點的就用最小路徑覆蓋好了,水菜傷不起
**********************************************************************/
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;

const int maxn = 40*10+10;
int map[45][15];

int g[maxn][maxn];
int linker[maxn];
int vis[maxn];
int uN,vN;


bool dfs(int u)
{
    for(int v = 1; v <= vN; v++)
    {
        if(!vis[v] && g[u][v])
        {
            vis[v] = 1;
            if(linker[v] == -1 || dfs(linker[v]))
            {
                linker[v] = u;
                return true;
            }
        }
    }
    return false;
}

int hungary()
{
    int sum = 0;
    memset(linker,-1,sizeof(linker));
    for(int u = 1; u <= uN; u++)
    {
        memset(vis,0,sizeof(vis));
        if(dfs(u))
            sum++;
    }
    return sum;
}

int main()
{
    int T;
    int row,col;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d%*c", &row,&col);

        char c;
        int num = 0; //記錄有多少個點
        memset(map,0,sizeof(map));
        memset(g,0,sizeof(g));
        for(int i = 1; i <= row; i++) //從 1 開始, map[][]周圍加邊
        {
            for(int j = 1; j <= col; j++)
            {
                scanf("%c", &c);
                if(c == '*') map[i][j] = ++num;
            }
            getchar();
        }
        uN = vN = num;

        for(int i = 1; i <= row; i++) //建圖
        {
            for(int j = 1; j <= col; j++)
            {
                if(map[i][j])
                {
                    if(map[i][j+1])
                        g[map[i][j]][map[i][j+1]] = 1;
                    if(map[i][j-1])
                        g[map[i][j]][map[i][j-1]] = 1;
                    if(map[i+1][j])
                        g[map[i][j]][map[i+1][j]] = 1;
                    if(map[i-1][j])
                        g[map[i][j]][map[i-1][j]] = 1;
                }
            }
        }
        printf("%d\n", num-hungary()/2); //最小路徑覆蓋數
    }
    return 0;
}