HDU 1045 Fire Net(dfs,跟8皇后问题很相似)
传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1045
Fire Net
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14670 Accepted Submission(s): 8861
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.
.X..
....
XX..
....
2
XX
.X
3
.X.
X.X
.X.
3
...
.XX
.XX
4
....
....
....
....
0
1
5
2
4
就是给你一个地图,地图上有一些墙,向地图内放炮台,不要让两个炮台能互相射击到对方,
当然中间有墙就可以阻隔他们。求最多能放多少炮台。
分析:
如果放炮台,在该点设置一个炮台标记,
判断某一点是否可以放炮台,就是从该点向四个方向找,碰到墙停止,碰到别的炮台就返回0,表示该点不能放炮台。
code:
#include <stdio.h>
#include <iostream>
#include <stdlib.h>
#include <algorithm>
#include <string.h>
using namespace std;
char G[][];
int n,sum;
bool check(int x,int y)
{
if(G[x][y]=='X')//该点是墙
return false;
for(int i=x;i<n;i++)//下
{
if(G[i][y]=='X')
break;
if(G[i][y]=='S')
return false;
}
for(int i=x;i>=;i--)//上
{
if(G[i][y]=='X')
break;
if(G[i][y]=='S')
return false;
}
for(int j=y;j<n;j++)//右
{
if(G[x][j]=='X')
break;
if(G[x][j]=='S')
return false;
}
for(int j=y;j>=;j--)//左
{
if(G[x][j]=='X')
break;
if(G[x][j]=='S')
return false;
}
return true;
} void dfs(int x,int y,int num)
{
if(x==n)
{
sum=max(sum,num);
return ;
}
for(int j=y;j<n;j++)
{
if(check(x,j))
{
G[x][j]='S';
dfs(x,j,num+);
G[x][j]='.';
}
}
dfs(x+,,num);
}
int main()
{
while(cin>>n,n)
{
for(int i=;i<n;i++)
{
for(int j=;j<n;j++)
{
cin>>G[i][j];
}
}
sum=;
dfs(,,);
cout<<sum<<endl;
}
return ;
}
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