Fire Net

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. 

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

5
1
5
2
4

题意：给定一个最大4*4的方形地图，里面有墙(X)和空地(.)。在每个空地上可以放大炮，但两个大炮如果在同一行或同一列并且之间没有墙阻隔的话，会互相攻击，所以不能同时存在。问最多能放多少个大炮。

题解：二分匹配。建图的话就是同一行连续的‘.‘看做为一个节点，为A集合，同一列也一样，为B集合。

        A，B中只要有交点就加边。

ACcode：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <queue>
#include <cmath>
#include <string>
#define  N 101010
#define MaxN 10100
#define INF  1<<29

using namespace std;

int uN,vN;
int linker[20];
char mp[6][6];
bool used[20];
int mp1[6][6],mp2[6][6],maze[6][6];
int n,m;

bool dfs(int u) {
    int v;
    for(v=1; v<=vN; v++)
        if(maze[u][v]&&!used[v]) {
            used[v]=true;
            if(linker[v]==-1||dfs(linker[v])) {
                linker[v]=u;
                return true;
            }
        }
    return false;
}

int hungary() {
    int res=0;
    int u;
    memset(linker,-1,sizeof(linker));
    for(u=1; u<=uN; u++) {
        memset(used,0,sizeof(used));
        if(dfs(u))  res++;
    }
    return res;
}

int main() {
    //freopen("in.txt","r",stdin);
    while(~scanf("%d",&n)&&n) {
        for(int i=0; i<n; i++) {
            scanf("%s",mp[i]);
        }
        int l1=1,l2=1;
        memset(mp1,0,sizeof mp1);
        memset(mp2,0,sizeof mp2);
        for(int i=0; i<n; i++) {
            for(int j=0; j<n; j++) {
                if(mp1[i][j])continue;
                if(mp[i][j]=='X') {
                    mp1[i][j]=0;
                    continue;
                }
                while(mp[i][j]=='.'&&j<n)mp1[i][j]=l1,j++;
                l1++;
            }
            for(int j=0; j<n; j++) {
                if(mp2[j][i])continue;
                if(mp[j][i]=='X') {
                    mp2[j][i]=0;
                    continue;
                }
                while(mp[j][i]=='.'&&j<n)mp2[j][i]=l2,j++;
                l2++;
            }
        }
        memset(maze,0,sizeof maze);
        for(int i=0; i<n; i++) {
            for(int j=0; j<n; j++) {
                if(mp[i][j]!='X') {
                    maze[mp1[i][j]][mp2[i][j]]=1;
                }
            }
        }
        uN=l1-1;
        vN=l2-1;
        int ans=hungary();
        printf("%d\n",ans);
    }
    return 0;
}