hdu1045 Fire Net(最大匹配+建模)

Fire Net

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12998    Accepted Submission(s): 7884

Problem Description

Suppose that we have a square city with straight streets. A map of a city is a square board with n rows and n columns, each representing a street or a piece of wall. 

A blockhouse is a small castle that has four openings through which to shoot. The four openings are facing North, East, South, and West, respectively. There will be one machine gun shooting through each opening. 

Here we assume that a bullet is so powerful that it can run across any distance and destroy a blockhouse on its way. On the other hand, a wall is so strongly built that can stop the bullets. 

The goal is to place as many blockhouses in a city as possible so that no two can destroy each other. A configuration of blockhouses is legal provided that no two blockhouses are on the same horizontal row or vertical column in a map unless there is at least one wall separating them. In this problem we will consider small square cities (at most 4x4) that contain walls through which bullets cannot run through. 

The following image shows five pictures of the same board. The first picture is the empty board, the second and third pictures show legal configurations, and the fourth and fifth pictures show illegal configurations. For this board, the maximum number of blockhouses in a legal configuration is 5; the second picture shows one way to do it, but there are several other ways. 



Your task is to write a program that, given a description of a map, calculates the maximum number of blockhouses that can be placed in the city in a legal configuration. 

 

Input

The input file contains one or more map descriptions, followed by a line containing the number 0 that signals the end of the file. Each map description begins with a line containing a positive integer n that is the size of the city; n will be at most 4. The next n lines each describe one row of the map, with a '.' indicating an open space and an uppercase 'X' indicating a wall. There are no spaces in the input file. 

 

Output

For each test case, output one line containing the maximum number of blockhouses that can be placed in the city in a legal configuration.

 

Sample Input

4.X......XX......2XX.X3.X.X.X.X.3....XX.XX4................0

 

Sample Output

51524

题意：黑点是一个攻击装置，攻击上下左右，方块是墙可以阻挡攻击，问在装置不能相互攻击的前提下，最多可以放置多少个装置

我们可以将边界与墙之前抽象为一个集合，因为这之间只能放一个装置；

用这种方法分别将行和列抽象，此时，每个集合的编号都得出，那么(i,j)这点就可以建图为    i这点的行集合编号——>j这点的列集合编号

然后求行集合与列集合之间的最大匹配即可

#include<bits/stdc++.h>#define mem(a,b) memset(a,b,sizeof(a))using namespace std;char e[10][10];int er[10][10],ec[10][10];int g[10][10];int match[10];bool vis[10];int n1,n2;bool dfs(int u){    for(int i=1;i<=n2;i++)    {        if(!vis[i]&&g[u][i])        {            vis[i]=true;            if(match[i]==-1||dfs(match[i]))            {                match[i]=u;                return true;            }        }    }    return false;}int main(){    int n;    while(scanf("%d",&n)&&n)    {        mem(match,-1);        mem(g,0);        mem(ec,0);        mem(er,0);        for(int i=1;i<=n;i++)        {            scanf("%s",e[i]+1);        }        n1=0,n2=0;        int p1=0,p2=0;        //获得行集合的编号        for(int i=1;i<=n;i++)        {            for(int j=1;j<=n;j++)            {                while(e[i][j]=='X'&&j<=n)                {                    j++;                }                p1++;                while(e[i][j]!='X'&&j<=n)                {                    er[i][j]=p1;                    n2=n2>p1?n2:p1;                    j++;                }            }        }        //获得列集合的编号        for(int i=1;i<=n;i++)        {            for(int j=1;j<=n;j++)            {                while(e[j][i]=='X'&&j<=n)                {                    j++;                }                p2++;                while(e[j][i]!='X'&&j<=n)                {                    ec[j][i]=p2;                    n1=n1>p2?n1:p2;                    j++;                }            }        }        //建图        for(int i=1;i<=n;i++)        {            for(int j=1;j<=n;j++)            {                if(e[i][j]!='X')                {                    g[ec[i][j]][er[i][j]]=1;                }            }        }        int ans=0;        for(int i=1;i<=n1;i++)        {            mem(vis,0);            if(dfs(i))ans++;        }            printf("%d\n",ans);    }    return 0;}