HDU 1045 Fire Net 二分图最大匹配

╰半夏微凉° 2022-05-23 11:46 142阅读 0赞

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.  
  
![2833d81666db37f9dec4df3db9d38734_v_1528093284][]  
  
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.

InputThe 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.  
OutputFor 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..
    ....
    2
    XX
    .X
    3
    .X.
    X.X
    .X.
    3
    ...
    .XX
    .XX
    4
    ....
    ....
    ....
    ....
    0

Sample Output

5
    1
    5
    2
    4

因为是在二分图训练里做的，所以理所当然想到二分图啦~一开始还以为是和棋盘那题一样的建图方式，后来发现棋盘X不可以走，而这道题X作为墙，可以使得同行/列的可行，所以建图稍微麻烦了一点。

#include<stdio.h>
    #include<string.h>
    #include<math.h>
    #include<iostream>
    #include<string>
    #include<algorithm>
    #include<map>
    #include<queue>
    #include<vector>
    using namespace std;
    int n,ma[205][205],vis[205],dx[205],l1,l2;
    char c[205][205];
    void bulid()
    {
    	int i,j,k;
    	l1=l2=1;
    	bool visit[205][205];
    	int a[205][205],b[205][205];
    	memset(a,0,sizeof(a));
    	memset(b,0,sizeof(b));
    	memset(visit,false,sizeof(visit));
    	for(i=1;i<=n;i++)
    	{
    		for(j=1;j<=n;j++)
    		{
    			if(!visit[i][j]&&c[i][j]!='X')
    			{
    				k=j;
    				while(k<=n&&c[i][k]!='X')  
                    {  
                        a[i][k]=l1;  
                        visit[i][k]=true;  
                        k++;  
                    }  
                    l1++;  
    			}
    		}
    	}
    	memset(visit,false,sizeof(visit));
    	for(i=1;i<=n;i++)
    	{
    		for(j=1;j<=n;j++)
    		{
    			if(!visit[i][j]&&c[i][j]!='X')
    			{
    				k=i;
    				while(k<=n&&c[k][j]!='X')  
                    {  
                        b[k][j]=l2;  
                        visit[k][j]=true;  
                        k++;  
                    }  
                    l2++;  
    			}
    		}
    	}
    	for(i=1;i<=n;i++)
    	{
    		for(j=1;j<=n;j++)
            {  
            	//printf("a:%d b:%d\n",a[i][j],b[i][j]);
                ma[a[i][j]][b[i][j]]=1;  
            }  
        }  
    } 
    int find(int i)
    {
    	for(int j=1;j<=l2;j++)
    	{
    		if(vis[j]==0&&ma[i][j])
    		{
    			vis[j]=1;
    			if(dx[j]==-1||find(dx[j]))
    			{
    				dx[j]=i;
    				return 1;
    			}
    		}
    	}
    	return 0;
    }
    int main()
    {
    	int i,j;	
    	while(scanf("%d",&n)!=EOF)
    	{
    		memset(ma,0,sizeof ma);
    		memset(dx,-1,sizeof dx);
    		if(n==0)break;
    		for(i=1;i<=n;i++)
    		{getchar();
    			for(j=1;j<=n;j++)
    			{
    				scanf("%c",&c[i][j]);
    			}
    		}
    		bulid();//建图 
    		int s=0;
    		for(i=1;i<=l1;i++)
    		{
    			memset(vis,0,sizeof vis);
    			s+=find(i);
    		}
    		printf("%d\n",s);
    	}
    }

然后看题解的时候发现直接深搜也可以~

#include<stdio.h>  
    #include<string.h>  
    #define N 110  
    char ma[N][N];  
    int n,ans;  
    int check(int r, int c)
    {  
        int flag=1,i;  
        for(i=r;i>=0;i--)
    	{ 
            if(ma[i][c]=='X') break;  
            if(ma[i][c]=='b')
    		{  
                flag=0;  
                break;  
            }  
        }  
        for(i=c;i>=0;i--)
    	{  
            if(ma[r][i]=='X') break;  
            if(ma[r][i]=='b')
    		{  
                flag=0;  
                break;  
            }  
        }  
        return flag;  
    }  
    void dfs(int pos,int sum)
    {  
        int r=pos/n,c=pos%n;  
        if(pos==n*n)
    	{  
            if(sum>ans) 
    		ans=sum;  
            return;  
        }  
        if(ma[r][c]=='.')
    	{  
            if(check(r,c)==1)
    		{  
                ma[r][c]='b'; 
                dfs(pos+1,sum+1);  
                ma[r][c]='.';  
            }  
        }   
        dfs(pos+1,sum);  
    }  
    int main()
    {  
        int i;  
        while(scanf("%d",&n))
    	{  
    		if(n==0)break;
            memset(ma,0,sizeof(ma));  
            for(i=0,ans=0;i<n;i++)
            scanf("%s",ma[i]);  
            dfs(0,0);  
            printf("%d\n",ans);  
        }  
        return 0;  
    }

[2833d81666db37f9dec4df3db9d38734_v_1528093284]: https://odzkskevi.qnssl.com/2833d81666db37f9dec4df3db9d38734?v=1528093284