hdu 1045 Fire Net(最小覆盖点+构图（缩点））

http://acm.hdu.edu.cn/showproblem.php?pid=1045

Fire Net

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1045

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

这题二分匹配， 也可以直接用DFS搜索，这里上二分匹配的代码

该题与poj2226做法相同，具体如何构图在本博客poj2226这道题的题解上讲解的很清楚，在这里就不再过多解释

链接如下：

http://www.cnblogs.com/qq2424260747/p/4717142.html

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#define N 310

using namespace std;

char maps[N][N];
int G[N][N], vis[N], used[N];
int n, x, y, a[N][N], b[N][N];

bool Find(int u)
{
    int i;
    for(i = 1 ; i <= y ; i++)
    {
        if(!vis[i] && G[u][i])
        {
            vis[i] = 1;
            if(!used[i] || Find(used[i]))
            {
                used[i] = u;
                return true;
            }
        }
    }
    return false;
}

void Build()
{
    int i, j;
    x = y = 0;
    memset(a, 0, sizeof(a));
    memset(b, 0, sizeof(b));
    for(i = 1 ; i <= n ; i++)
    {
        for(j = 1 ; j <= n ; j++)
        {
            if(maps[i][j] == ‘.‘)
            {
                if(maps[i][j - 1] == ‘.‘)
                    a[i][j] = a[i][j - 1];
                else
                    a[i][j] = ++x;
            }
        }
    }
    for(i = 1 ; i <= n ; i++)
    {
        for(j = 1 ; j <= n ; j++)
        {
            if(maps[i][j] == ‘.‘)
            {
                if(maps[i - 1][j] == ‘.‘)
                    b[i][j] = b[i - 1][j];
                else
                    b[i][j] = ++y;
                G[a[i][j]][b[i][j]] = 1;
            }
        }
    }

}

int main()
{
    int i, j, ans;
    while(scanf("%d", &n), n)
    {
        ans = 0;
        memset(G, 0, sizeof(G));
        for(i = 1 ; i <= n ; i++)
        {
            getchar();
            for(j = 1 ; j <= n ; j++)
                 scanf("%c", &maps[i][j]);
        }
        Build();
        memset(used, 0, sizeof(used));
        for(i = 1 ; i <= x ; i++)
        {
            memset(vis, 0, sizeof(vis));
            if(Find(i))
                ans++;
        }
        printf("%d\n", ans);
    }
    return 0;
}