HDU-1045 Fire Net

浏览数：14 / 时间：2015年06月09日

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7123 Accepted Submission(s): 4057

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.

技术分享

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

Source

Zhejiang University Local Contest 2001

Recommend

/**       
          刚看题，没有想到二分图匹配，然后查了题解，
          题意：是在一个4*4的矩阵中放大炮，看能放大炮的最大数，(.)表示空地，(X) 表示墙，
                    如果大炮所在的该行或该列存在大炮，然后就不能相互攻击，否则能够相互攻击
          做法：二分图最大匹配 匈牙利算法
          
**/
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<cmath>
#include<string.h>
using namespace std;
#define maxn 1100
bool  g[maxn][maxn];
bool used[maxn];
int un,vn;
int linker[maxn];
int mmap[maxn];
int row[maxn][maxn];
 int roww = 0,coll = 0;
int col[maxn][maxn];
int  dfs(int u)
{
    for(int i =0 ; i<coll; i++)
    {
        if(g[u][i] &&used[i]==false)
        {
            used[i]  =true;
            if(linker[i] == -1 || dfs(linker[i]))
            {
                linker[i] = u;
                mmap[u] = i;
                return 1;
            }
        }
    }
    return 0;
}
int hungry()
{
    int res = 0;
    memset(linker,-1,sizeof(linker));
    memset(mmap,-1,sizeof(mmap));
    for(int u = 0; u<roww; u++)
    {
        if(mmap[u] == -1)
        {
            memset(used,false,sizeof(used));
            res += dfs(u);
        }
    }
    return res;
}
char ch[110][110];
int main()
{
#ifndef ONLINE_JUDGE
          freopen("in.txt","r",stdin);
#endif // ONLINE_JUDGE
    int n;
    while(~scanf("%d",&n))
    {
        if(n == 0) break;
        for(int i=0; i<n; i++)
        {
            scanf("%s",ch[i]);
        }
                    memset(row,-1,sizeof(row));
                    memset(col,-1,sizeof(col));
                    roww = 0;
                    coll = 0;
                    for(int i=0;i<n;i++)
                    {
                              for(int j=0;j<n;j++)
                              {
                                        if(ch[i][j] == ‘.‘ && row[i][j] == -1)
                                        {
                                                  for(int k=j;ch[i][k] == ‘.‘ &&k<n;k++)
                                                  {
                                                            row[i][k] = roww;
                                                  }
                                                  roww++;
                                        }
                                        if(ch[j][i] == ‘.‘&& col[j][i] == -1)
                                        {
                                                  for(int k = j;ch[k][i] == ‘.‘ && k<n;k++)
                                                  {
                                                            col[k][i] = coll;
                                                  }
                                                  coll++;
                                        }
                              }
                    }
                    memset(g,false,sizeof(g));
                    for(int i=0;i<n;i++)
                    {
                              for(int j=0;j<n;j++)
                              {
                                        if(ch[i][j]== ‘.‘) g[row[i][j]][col[i][j]] = true;
                              }
                    }
                   printf("%d\n", hungry());
    }
    return 0;
}