YT04-貪心課堂練習-1004—Fire Net-(6.14日-煙臺大學ACM預備隊解題報告)
Fire Net
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 9 Accepted Submission(s) : 7
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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
Source
Zhejiang University Local Contest 2001
bool allow(int n, int row, int col)
{
if(net[row][col] == 'X')
{
return false;
}
int i;
for(i = row; i >= 0; --i) //向前遍歷該列
{
if(net[i][col] == '@') //若該列防止過碉堡
return false; //則該位置不能放置
else if(net[i][col] == 'X') //若遇到牆
break; //跳出迴圈遍歷開始行
}
for(i = col; i >= 0; --i) //向前遍歷該行
{
if(net[row][i] == '@') //同上
return false;
else if(net[row][i] == 'X')
break;
}
return true;
}
bool allow(int n, int row, int col)
{
if(net[row][col] == 'X')
{
return false;
}
int i;
for(i = row; i >= 0; --i) //向前遍歷該列
{
if(net[i][col] == '@') //若該列防止過碉堡
return false; //則該位置不能放置
else if(net[i][col] == 'X') //若遇到牆
break; //跳出迴圈遍歷開始行
}
for(i = col; i >= 0; --i) //向前遍歷該行
{
if(net[row][i] == '@') //同上
return false;
else if(net[row][i] == 'X')
break;
}
return true;
}
void BackTrack(int n, int k, int num)
{
if(k == n * n)
{
if(num > max)
{
max = num;
}
return;
}
int row = k / n, col = k % n;
if(allow(n, row, col))
{
net[row][col] = '@';
BackTrack(n, k + 1, num + 1);
net[row][col] = '.';
}
BackTrack(n, k + 1, num);
}
int main()
{
int n, i;
while(scanf("%d", &n) && n)
{
max = 0;
for(i = 0; i < n; ++i)
{
scanf("%s", net[i]);
}
BackTrack(n, 0, 0);
printf("%d\n", max);
}
return 0;
}
回溯演算法
•回溯演算法也叫試探法,它是一種系統地搜尋問題的解的方法。回溯演算法的基本思想是:從一條路往前走,能進則進,不能進則退回來,換一條路再試。用回溯演算法解決問題的一般步驟為:
•1、定義一個解空間,它包含問題的解。
•2、利用適於搜尋的方法組織解空間。
•3、利用深度優先法搜尋解空間。
•4、利用限界函式避免移動到不可能產生解的子空間。
•問題的解空間通常是在搜尋問題的解的過程中動態產生的,這是回溯演算法的一個重要特性。
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