ZOJ 1002 Fire Net

                                                                           Fire  Net


                                                 Time Limit: 1 Second Memory Limit: 32768 KB

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.

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.

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

#include<stdio.h>
#define N 200

 

int c,bestc;
int n;
char a[N][N];

 

int testa(int x,int y,int z)
{
 for(int i=y+1;i<z;i++)
  if(a[x][i]=='X') return 1;
 return 0;
   
}

 

int testb(int x,int y,int z)
{
 for(int i=y+1;i<z;i++)
  if(a[i][x]=='X') return 1;
 return 0;
}

 

int test(int r,int l)
{
 for(int i=0;i<l;i++)
  if(a[r][i]=='A'&&!testa(r,i,l))
   return 0;
 for(int j=0;j<r;j++)
  if(a[j][l]=='A'&&!testb(l,j,r))
   return 0;
 return 1; 
}

 

void backtrack(int t)
{
 int row,col;
 row=t/n;
 col=t%n;

 if(t>=n*n)
 {
  if(c>bestc)
   bestc=c; 
 } 
 else
 {  
  if(a[row][col]=='.'&&test(row,col))
  {
   a[row][col]='A';
   c++;
 
   backtrack(t+1);
   a[row][col]='.';
   c--; 
  }
  backtrack(t+1); 
 }
}

 

int main()
{
 while( scanf("%d",&n), n!= 0 )
 {
  getchar();
  for(int i=0;i<n;i++)
  {
   for(int j=0;j<n;j++)
    scanf("%c",&a[i][j]);
   getchar();
     }
     bestc=c=0;
  backtrack(0);
  printf("%d/n",bestc);
 }
 return 0; 
}

                           

 

基于python实现的粒子群的VRP(车辆配送路径规划)问题建模求解+源码+项目文档+算法解析,适合毕业设计、课程设计、项目开发。项目源码已经过严格测试,可以放心参考并在此基础上延申使用,详情见md文档 算法设计的关键在于如何向表现较好的个体学习,标准粒子群算法引入惯性因子w、自我认知因子c1、社会认知因子c2分别作为自身、当代最优解和历史最优解的权重,指导粒子速度和位置的更新,这在求解函数极值问题时比较容易实现,而在VRP问题上,速度位置的更新则难以直接采用加权的方式进行,一个常见的方法是采用基于遗传算法交叉算子的混合型粒子群算法进行求解,这里采用顺序交叉算子,对惯性因子w、自我认知因子c1、社会认知因子c2则以w/(w+c1+c2),c1/(w+c1+c2),c2/(w+c1+c2)的概率接受粒子本身、当前最优解、全局最优解交叉的父代之一(即按概率选择其中一个作为父代,不加权)。 算法设计的关键在于如何向表现较好的个体学习,标准粒子群算法引入惯性因子w、自我认知因子c1、社会认知因子c2分别作为自身、当代最优解和历史最优解的权重,指导粒子速度和位置的更新,这在求解函数极值问题时比较容易实现,而在VRP问题上,速度位置的更新则难以直接采用加权的方式进行,一个常见的方法是采用基于遗传算法交叉算子的混合型粒子群算法进行求解,这里采用顺序交叉算子,对惯性因子w、自我认知因子c1、社会认知因子c2则以w/(w+c1+c2),c1/(w+c1+c2),c2/(w+c1+c2)的概率接受粒子本身、当前最优解、全局最优解交叉的父代之一(即按概率选择其中一个作为父代,不加权)。
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