Fire Net
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 1 Accepted Submission(s) : 1
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Problem Description
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 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
Sample Input
4
.X..
....
XX..
....
2
XX
.X
3
.X.
X.X
.X.
3
...
.XX
.XX
4
....
....
....
....
0
/*题解：
DFS+回溯，对于图上的每个点，只有能放大炮和不能放两种情况，若能放大炮则有放或者不放两种情况，若不能放，则只有一种。
很经典的处理方法。DFS初体验，感觉很不错。 参考自：http://blog.csdn.net/u013365671/article/details/38292825
*/
#include<cstdio>
#include<cstring>
char a[8][8];
int ans,n;
int judge(int p)//判断是否可以放置大炮
{
int x=p/n,y=p%n;//x行，y列
for(int i=x; i>=0; i--)//向上搜
{
if(a[i][y]=='#') return 0;//'#'代表大炮
if(a[i][y]=='X') break;
}
for(int i=y; i>=0; i--)//向左搜
{
if(a[x][i]=='#') return 0;
if(a[x][i]=='X') break;
}
return 1;
}
void DFS(int p,int count)//深搜
{
if(p==n*n)
{
if(count>ans)
ans = count;
return;
}
int x=p/n,y=p%n;
if(judge(p)&&a[x][y]=='.')
{
a[x][y] = '#';//放大炮
DFS(p+1,count+1);
a[x][y] = '.';//回溯为原来状态
}
DFS(p+1,count);//不放大炮
}
int main()
{
while(scanf("%d",&n),n)
{
memset(a,0,sizeof(a));
ans=0;
for(int i=0; i<n; i++)
scanf("%s",a[i]);
DFS(0,0);
printf("%d\n",ans);
}
return 0;
}
注释：深度优先搜索属于图算法的一种,英文缩写为DFS即Depth First Search.
其过程简要来说是对每一个可能的分支路径深入到不能再深入为止,
而且每个节点只能访问一次.