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.

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.

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

ZOJ 1002
摆碉堡，碉堡之间不能互相攻击

  1 #include <stdio.h>
  2 #include <stdlib.h>
  3 #include <string.h>
  4 struct linjiejuzhen
  5 {
  6     int xiabiao;
  7     int keyigongcun[16];
  8 };
  9 void inpt();//输入数据
 10 void creat(struct linjiejuzhen *p);//建立互相之间能不能打到的邻接矩阵
 11 void read(struct linjiejuzhen *p,int mark[],int num);//递归地搜索互相之间不会攻击的碉堡
 12 struct linjiejuzhen head[16];
 13 char a[17];//地形图，行数由n记录
 14 int n;//长宽
 15 int longest;//最大值
 16 int main()
 17 {
 18     int i,j;
 19     int mark[16];
 20     do
 21     {
 22         scanf("%d",&n);
 23         if(n==0)
 24         {
 25             break;
 26         }
 27         longest=0;
 28         inpt();
 29         for(i=0;i<n*n;i++)
 30         {
 31             head[i].xiabiao=i;
 32         }
 33         for(i=0;i<n*n;i++)
 34         {
 35             creat(&head[i]);
 36         }
 37         for(i=0;i<n*n;i++)
 38         {
 39             for(j=0;j<n*n;j++)
 40             {
 41                 mark[j]=1;
 42             }
 43             if(a[i]!='X')
 44                 read(&head[i],mark,1);
 45         }
 46         printf("%d\n",longest);
 47     }while(n!=0);
 48     return 0;
 49 }
 50 void inpt()
 51 {
 52     char b[5];
 53     int i;
 54     for(i=0;i<17;i++)
 55     {
 56         a[i]='\0';
 57     }
 58     for(i=0;i<n;i++)
 59     {
 60         scanf("%s",b);
 61         strcat(a,b);
 62     }
 63 }
 64 void creat(struct linjiejuzhen *p)
 65 {
 66     int i,j;
 67     int x;
 68     x=p->xiabiao;
 69     for(i=0;i<n*n;i++)
 70     {
 71         p->keyigongcun[i]=0;
 72     }
 73     if(a[x]!='X')
 74     {
 75         for(j=0;j<n*n;j++)
 76         {
 77             if(j!=x&&a[j]!='X')
 78             {
 79                 p->keyigongcun[j]=1;
 80             }
 81         }
 82     }
 83     i=x-1;
 84     while(i>=(x/n)*n)
 85     {
 86         if(a[i]=='.')
 87         {
 88             p->keyigongcun[i]=0;
 89         }
 90         else if(a[i]=='X')
 91         {
 92             break;
 93         }
 94         i--;
 95     }
 96     i=x+1;
 97     while(i<(x/n+1)*n)
 98     {
 99         if(a[i]=='.')
100         {
101             p->keyigongcun[i]=0;
102         }
103         else if(a[i]=='X')
104         {
105             break;
106         }
107         i++;
108     }
109     i=x-n;
110     while(i>=0)
111     {
112         if(a[i]=='.')
113         {
114             p->keyigongcun[i]=0;
115         }
116         else if(a[i]=='X')
117         {
118             break;
119         }
120         i-=n;
121     }
122     i=x+n;
123     while(i<n*n)
124     {
125         if(a[i]=='.')
126         {
127             p->keyigongcun[i]=0;
128         }
129         else if(a[i]=='X')
130         {
131             break;
132         }
133         i+=n;
134     }
135 }
136 /*搜索可以共存的碉堡
137   通过记录每一个
138   已经记录的碉堡的不能共存的碉堡*/
139 void read(struct linjiejuzhen *p,int mark[],int num)
140 {
141     int ed[16];
142     int i;
143     int flag=0;
144     for(i=0;i<n*n;i++)
145     {
146         ed[i]=mark[i];
147     }
148     for(i=0;i<n*n;i++)
149     {
150         if(p->keyigongcun[i]==0)
151         {
152             ed[i]=0;
153         }
154     }
155     for(i=0;i<n*n;i++)
156     {
157         if(ed[i]==1)
158         {
159             flag=1;
160             break;
161         }
162     }
163     if(flag==0)
164     {
165         if(num>longest)
166         {
167             longest=num;
168         }
169     }
170     for(i=0;i<n*n;i++)
171     {
172         if(ed[i]==1)
173         {
174             read(&head[i],ed,num+1);
175         }
176     }
177 }

View Code

/*只学了C
  有很多地方可以优化
  感觉自己写的好繁琐*/
//数学模型没好好想，一开始搜索方法不对
/*既然互相不能攻击就是一个搜索分支的任何点都不能互相攻击
  那么应该每记录一点后都把新产生的攻击点“添加”进不能摆的点的记录中
  下次搜索时便不记录这个点*/

posted on 2019-11-21 21:41 KNGS 阅读(106) 评论(0) 收藏举报

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