Fire Net
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7929 Accepted Submission(s):
4521
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.
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
用深搜对每一个点进行遍历，
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 char map[5][5]; 6 int n,now,ibest,current; 7 bool canput(int row,int col)//判断能不能放 8 { 9 int i; 10 for(i=row-1;i>=0;i--)//判断行 11 { 12 if(map[i][col]=='O') 13 return false; 14 if(map[i][col]=='X') 15 break; 16 } 17 for(i=col-1;i>=0;i--)//判断列 18 { 19 if(map[row][i]=='O') 20 return false; 21 if(map[row][i]=='X') 22 break; 23 } 24 return true; 25 } 26 void dfs(int k,int current) 27 { 28 int x,y; 29 if(k==n*n)//如果遍历完就返回 30 { 31 if(current>ibest)//更新最大的个数 32 ibest=current; 33 return ; 34 } 35 else 36 { 37 x=k/n; 38 y=k%n; 39 if(map[x][y]=='.'&&canput(x,y)) 40 { 41 map[x][y]='O'; 42 dfs(k+1,current+1);//下一次递归 43 map[x][y]='.'; 44 } 45 dfs(k+1,current);//当前不放碉堡 46 } 47 } 48 int main() 49 { 50 while(scanf("%d",&n),n) 51 { 52 getchar(); 53 int k=0,i; 54 ibest=0; 55 current=0; 56 for(i=0;i<n;i++) 57 scanf("%s",map[i]); 58 dfs(0,0); 59 printf("%d ",ibest); 60 } 61 return 0; 62 }