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
题意：有n条n列的街道，街道上‘.‘表示空地，‘X‘表示障碍物，现在要求在街道上建立房子，但是每行每列如果没有障碍物，则只能有一间房子，如果有障碍物阻拦则可能不止可以建一座房子，问最多可以建几座房子
1.二分匹配：
定义一段无法向左右扩展的连续非障碍物方格为行连通块，一段无法向上下扩展的连续非障碍物方格为列连通块
我们把行连通块对应X集合，列连通块对应Y集合，如果一个行连通块与一个列连通块有交点且为空地，则对应在二分图中有一条边
显然房子的集合对应二分图的一个匹配，任意两个房子不可能共同存在于一个行或列连通块中，所以最多房子数目=最大匹配数
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define N 10
using namespace std;
char Map[N][N];
int G[N][N];
int use[N], vis[N];
int n, rn, cn;
int Find(int u)
{
int i;
for (i = 1; i <= cn; i++)
{
if (!vis[i] && G[u][i])
{
vis[i] = 1;
if (!use[i] || Find(use[i]))
{
use[i] = u;
return 1;
}
}
}
return 0;
}
int main ()
{
int i, j, r, c, ans;
int mapr[N][N], mapc[N][N];
while (scanf("%d", &n), n)
{
r = c = rn = cn = ans = 0;
memset(G, 0, sizeof(G));
memset(use, 0, sizeof(use));
memset(mapr, 0, sizeof(mapr));
memset(mapc, 0, sizeof(mapc));
for (i = 0; i < n; i++)
scanf("%s", Map[i]);
for (i = 0; i < n; i++)
for (j = 0; j < n; j++)
if (Map[i][j] == ‘X‘)
mapr[i][j] = mapc[i][j] = -1;
for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
{
while (mapr[i][j] == -1 && j < n)
j++;
r++;
while (mapr[i][j] != -1 && j < n)
{
mapr[i][j] = r; //建立行连通块
if (rn < r) rn = r;
j++;
}
}
}
for (j = 0; j < n; j++)
{
for (i = 0; i < n; i++)
{
while (mapc[i][j] == -1 && i < n)
i++;
c++;
while (mapc[i][j] != -1 && i < n)
{
mapc[i][j] = c; //建立列连通块
if (cn < c) cn = c;
i++;
}
}
}
for (i = 0; i < n; i++)
for (j = 0; j < n; j++)
if (mapr[i][j] != -1 && mapc[i][j] != -1)
G[mapr[i][j]][mapc[i][j]] = 1; //重新构图
for (i = 1; i <= rn; i++)
{
memset(vis, 0, sizeof(vis));
if (Find(i)) ans++;
}
printf("%d\n", ans);
}
return 0;
}
2.该题还可以暴力搜索：
#include<stdio.h>
#include<algorithm>
#define N 10
using namespace std;
char Map[N][N];
int n, Max;
int Judge(int x, int y) //判断该行该列是否已经建过房子或者有障碍物
{
int i;
for (i = x-1; i >= 0; i--)
{
if (Map[i][y] == ‘#‘) //建过房子，则不能再建
return 0;
if (Map[i][y] == ‘X‘) //有障碍物，则还能再建
break;
}
for (i = y-1; i >= 0; i--)
{
if (Map[x][i] == ‘#‘)
return 0;
if (Map[x][i] == ‘X‘)
break;
}
return 1;
}
void DFS(int s, int c) //s表示走的步数，c表示可以建几座房子
{
int x, y;
Max = max(Max, c);
if (s == n*n) return ;
x = s / n; //x表示行
y = s % n; //y表示列
if (Map[x][y] == ‘.‘ && Judge(x, y))
{
Map[x][y] = ‘#‘;
DFS(s+1, c+1); //如果这点可以建立房子，则步数增加，房子总数增加
Map[x][y] = ‘.‘;
}
DFS(s+1, c); //即使这个点不能建房子，步数依然要增加
}
int main ()
{
int i;
while (scanf("%d", &n), n)
{
for (i = 0; i < n; i++)
scanf("%s", Map[i]);
Max = 0;
DFS(0, 0);
printf("%d\n", Max);
}
return 0;
}