HDU 1045 Fire Net

Fire Net
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
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

（Ps: my DFS 第一题）
题目大意：一个图上尽可能建多一点城堡，但不能使他们互相攻击（不能同一行或同一列上除非中间有墙）

思路：题目数据不大，暴力枚举，用DFS可以解决这个问题。

#include<iostream>
#include<cstdlib>
using namespace std;
#define MVNum 5 //最大顶点数
typedef char ArcType; 
struct AMGraph{
    ArcType arcs[MVNum][MVNum];//邻接矩阵
    int vexnum;
};
void CreateUDN(AMGraph &G)
{
    for(int i = 0; i < G.vexnum; i++)
    for(int j = 0; j < G.vexnum; j++)
        cin>>G.arcs[i][j];
}
int ans;
int dir[4][2] = {-1,0,1,0,0,-1,0,1}; //上，下，左，右 
void DFS_AM(AMGraph &G, int sum)
{//sum为当前炮台的数量 
    if(sum > ans)ans = sum;
    for(int i = 0; i < G.vexnum; i++)
    for(int j = 0; j < G.vexnum; j++)
    {
        if(G.arcs[i][j] == '.')
        {//空地，检查是否可以建立炮台 
            bool flag = true; //是否可建标志 
            for(int w = 0; w < 4; w++)//检查上下左右四个方向
            {
                int x = i, y = j;
                while(true)
                {
                    x = x + dir[w][0];
                    y = y + dir[w][1];
                    if(x < 0 || y < 0 || x == G.vexnum || y == G.vexnum)break; //到达边界
                    else if(G.arcs[x][y] == 'X') break;//遇到墙
                    else if(G.arcs[x][y] == 'P')//遇到炮台 
                    {
                        flag = false;
                        break;
                    }
                }
            }
            if(flag)
            {
                G.arcs[i][j] = 'P';
                DFS_AM(G, sum+1);//如果可建炮台，在当前图进行深搜 
                G.arcs[i][j] = '.';
            }
        }
    }
}
int main()
{
    ios::sync_with_stdio(false);
    AMGraph G;
    int n;
    while(cin>>n && n != 0)
    {
        G.vexnum = n;
        CreateUDN(G);
        ans = 0;
        DFS_AM(G, ans);
        cout<<ans<<endl;
    }
    return 0;
}