hdu 1045 二分匹配练习 重在转化啊！！

Fire Net

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10212    Accepted Submission(s): 5975

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..
....
2
XX
.X
3
.X.
X.X
.X.
3
...
.XX
.XX
4
....
....
....
....
0

 

Sample Output

5
1
5
2
4

 
忍不住再水一道     这是我第一次完全是自己转化出来的二分匹配题    先说好代码丑 臭 长    想看简介的还是点下标签的  叉叉  吧
但毕竟是自己的亲儿子   蛤蛤

因为这题的矩阵范围是4*4   特别小   暴搜也是随便过的    这里我就不贴了   和之前写的一道超暴力回溯差不多    但是简单的很多

还是讲讲二分的吧   如果是平时一对一的匹配   只要  把行看成一个集合  把列看成一个集合进行匹配即可    但这里一行可以与多列匹配   一列与多行匹配  需要重新建图

我的烂思路就是再开两个矩阵    一个记录新的  行   一个记录新的列   再建立方向矩阵   套模板即可

ac code

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;
int n,nrow,ncol;
const int maxn=16;
bool g[maxn][maxn],vis[maxn];
char mat[maxn][maxn];
int row[maxn][maxn],col[maxn][maxn],link[maxn];
int hungary();
int dfs(int x);

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0) break;
        getchar();
        for(int i=1; i<=n; i++) scanf("%s",mat[i]+1);
        int num=1,flag[3];
        for(int i=1; i<=n; i++)//建立行矩阵
        {
            flag[0]=flag[1]=flag[2]=0;
            for(int j=1; j<=n; j++)
                if(mat[i][j]=='X')
                {
                    row[i][j]=0;
                    flag[1]=j;
                }
                else
                {
                    if(flag[0]&&flag[1]&&flag[0]<flag[1])//如果同时存在.和X  并且  .  在X 前面
                    {
                        row[i][j]=n+num;
                        flag[2]=1;
                    }
                    else row[i][j]=i;
                    if(!flag[0]) flag[0]=j;
                }
            if(flag[2]) num++;
        }
        nrow=n+num-1;//新行长度

        num=1;
        for(int i=1; i<=n; i++)//建立新列 与上相同
        {
            flag[0]=flag[1]=flag[2]=0;
            for(int j=1; j<=n; j++)
                if(mat[j][i]=='X')
                {
                    col[j][i]=0;
                    flag[1]=j;
                }
                else
                {
                    if(flag[0]&&flag[1]&&flag[0]<flag[1])
                    {
                        col[j][i]=n+num;
                        flag[2]=1;
                    }
                    else col[j][i]=i;
                    if(!flag[0]) flag[0]=j;
                }
            if(flag[2]) num++;
        }
        ncol=n+num-1;

        memset(g,false,sizeof g);
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                g[row[i][j]][col[i][j]]=true;//建立邻接矩阵
        g[0][0]=false;
        int ans=hungary();//套模板
        printf("%d\n",ans);
    }
    return 0;
}

int hungary()
{
    int num=0;
    memset(link,-1,sizeof link);
    for(int i=1;i<=nrow;i++)
    {
        memset(vis,false,sizeof vis);
        num+=dfs(i);
    }
    return num;
}

int dfs(int x)
{
    for(int i=1;i<=ncol;i++)
        if(!vis[i]&&g[x][i])
        {
            vis[i]=true;
            if(link[i]==-1||dfs(link[i]))
            {
                link[i]=x;
                return 1;
            }
        }
    return 0;
}

学长总是说   图论只要套模板就好了   还有个关键就是你要套的来阿……

所以   以我现在的水平判断    转化建图再图论里是比较重要的          有些题    你根本看不出来是图论题    但是一转化  就会豁然开朗