zoukankan      html  css  js  c++  java
  • hdu 1045 Fire Net

    Fire Net

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 12661    Accepted Submission(s): 7666


    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.
     
    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
     
     
    dfs
     
     1 #include <iostream>
     2 #include <stdio.h>
     3 #include <algorithm>
     4 #include <cmath>
     5 #include <cstring>
     6 using namespace std;
     7 char mp[5][5];
     8 int ans,n,vis[5][5];
     9 int jiancha(int i,int j)
    10 {
    11     if(vis[i][j]) return 0;
    12     for(int di=i+1;di<n;di++)
    13     {
    14         if(vis[di][j]==1)
    15         return 0;
    16         if(vis[di][j]==2)
    17         break;
    18     }
    19     for(int di=i-1;di>=0;di--)
    20     {
    21         if(vis[di][j]==1)
    22         return 0;
    23         if(vis[di][j]==2)
    24         break;
    25     }
    26     for(int di=j-1;di>=0;di--)
    27     {
    28         if(vis[i][di]==1)
    29         return 0;
    30         if(vis[i][di]==2)
    31         break;
    32     }
    33     for(int di=j+1;di<n;di++)
    34     {
    35         if(vis[i][di]==1)
    36         return 0;
    37          if(vis[i][di]==2)
    38         break;
    39     }
    40     return 1;
    41 }
    42 void dfs(int m)
    43 {
    44     for(int i=0;i<n;i++)
    45     {
    46         for(int j=0;j<n;j++)
    47         {
    48             if(jiancha(i,j))
    49             {
    50                 vis[i][j]=1;
    51                 dfs(m+1);
    52                 vis[i][j]=0;
    53             }
    54         }
    55     }
    56     ans=max(ans,m);
    57 }
    58 
    59 int main()
    60 {
    61     while(~scanf("%d",&n)&&n)
    62     {
    63         memset(vis,0,sizeof(vis));
    64         for(int i=0; i<n; i++)
    65         {
    66             getchar();
    67             for(int j=0;j<n;j++)
    68             {
    69                 scanf("%c",&mp[i][j]);
    70                 if(mp[i][j]=='X')
    71                 vis[i][j]=2;
    72             }
    73         }
    74         ans=0;
    75         dfs(0);
    76         printf("%d
    ",ans);
    77     }
    78     return 0;
    79 }
    View Code
  • 相关阅读:
    js之oop <四>对象管理
    收集的一些前端面试题(持续更新)
    js之oop <三>属性标签
    js之oop <二> 对象属性
    js之oop <一> 创建对象,构造器(class)
    PHP_$_SERVER_说明详解
    利用jquery实现网站中对应栏目下面内容切换效果。
    网站中通知公告栏目滚动提示的效果
    图片上传利用<iframe></iframe>标签实现无刷新上传图片
    phpcms v9 数据库操作函数
  • 原文地址:https://www.cnblogs.com/52why/p/7483118.html
Copyright © 2011-2022 走看看