1045Fire Net
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题目大意:
在地图上,建碉堡,X表示墙,不能在X位置上建碉堡;. 表示可以建的地方
要求:在没有墙阻隔的情况下,两碉堡不能建在同一行或同一列,求最大建造的个数
解题思路:
利用回溯法求解,遍历整个图,得到建造数最多的情况
题目:
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......2XX.X3.X.X.X.X.3....XX.XX4................0
Sample Output
51524
代码:
#include<iostream>#include<String>#include<algorithm>using namespace std;char map[6][6];int n;int best;bool isplay(int x,int y){if(map[x][y]=='X')return false;for(int i=x-1;i>=0;i--){if(map[i][y]=='X')break;if(map[i][y]=='A') //碰到房子return false;}for(int j=y-1;j>=0;j--){if(map[x][j]=='X')break;if(map[x][j]=='A')return false;}return true;}void dfs(int p,int sum) //到达p点时,放置的房子数 {if(p==n*n){//搜索结束if(sum>best)best=sum;return;}int x=p/n;int y=p%n;if(isplay(x,y)){map[x][y]='A'; //建房子dfs(p+1,sum+1);map[x][y]='.';}dfs(p+1,sum);}int main(){while(cin>>n,n){int x=0,y=0;int sum=0;for(int i=0;i<n;i++)for(int j=0;j<n;j++)cin>>map[i][j];best=0;dfs(0,0);cout<<best<<endl;}return 0;}
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