2013
11-26

# Fire Net

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 4×4) 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.

4
.X..
....
XX..
....
2
XX
.X
3
.X.
X.X
.X.
3
...
.XX
.XX
4
....
....
....
....
0

5
1
5
2
4

#include <iostream>
using namespace std;

char map[10][10];
int n;
int ans;

int judge(int k) //判断该点能否放置大炮
{
int x = k / n;
int y = k % n;
int i;
for(i=y; i>=0; i--)		//从该点往左找
{
if(map[x][i] == 'X')
{
break;
}
if(map[x][i] == '@')
{
return false;
}
}
for(i=x; i>=0; i--)   //从该点往上找
{
if(map[i][y] == 'X')
{
break;
}
if(map[i][y] == '@')
{
return false;
}
}
return true;
}

void dfs(int k, int cnt)
{
if(k == n * n)
{
if(cnt > ans)
{
ans = cnt;
}
return ;
}
if(map[k/n][k%n] == '.' && judge(k))  //若该点能放置大炮
{
map[k/n][k%n] = '@';      //'@' 标记为大炮
dfs(k+1,cnt+1);
map[k/n][k%n] = '.';	//回溯时恢复该点状态
}
dfs(k+1,cnt);  //不放
}

int main()
{
while(cin>>n,n)
{
ans = 0;
for(int i=0; i<n; i++)
{
cin>>map[i];
}
dfs(0,0);
cout<<ans<<endl;
}
return 0;
}

1. 换句话说，A[k/2-1]不可能大于两数组合并之后的第k小值，所以我们可以将其抛弃。
应该是，不可能小于合并后的第K小值吧

2. 第一句可以忽略不计了吧。从第二句开始分析，说明这个花色下的所有牌都会在其它里面出现，那么还剩下♠️和♦️。第三句，可以排除2和7，因为在两种花色里有。现在是第四句，因为♠️还剩下多个，只有是♦️B才能知道答案。

3. #include <cstdio>

int main() {
//answer must be odd
int n, u, d;
while(scanf("%d%d%d",&n,&u,&d)==3 && n>0) {
if(n<=u) { puts("1"); continue; }
n-=u; u-=d; n+=u-1; n/=u;
n<<=1, ++n;
printf("%dn",n);
}
return 0;
}