Fire Net


Time Limit: 2 Seconds      Memory Limit: 65536 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

题意

大小为n*n的城市建造碉堡,要求碉堡建在‘.’位置,每两个碉堡不能在一行或一列,或者在一行一列的时候中间有‘X’隔开,问在这个城市中最多能建多少碉堡

思路

从左上角往右下角进行dfs,用一个check函数来判断当前位置是否可以建造碉堡,如果可以的话,将该位置做特殊标记。回溯寻找最大值

AC代码

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <limits.h>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <string>
#define ll long long
#define ull unsigned long long
#define ms(a) memset(a,0,sizeof(a))
#define pi acos(-1.0)
#define INF 0x7f7f7f7f
#define lson o<<1
#define rson o<<1|1
const double E=exp(1);
const int maxn=1e3+10;
const int mod=1e9+7;
using namespace std;
char ch[maxn][maxn];
int n;
int ans;
bool check(int x,int y)
{
for(int i=x-1;i>=0;i--)
{
if(ch[i][y]=='%')
return false;
if(ch[i][y]=='X')
break;
}
for(int i=y-1;i>=0;i--)
{
if(ch[x][i]=='%')
return false;
if(ch[x][i]=='X')
break;
}
return true;
}
void dfs(int s,int sum)
{
if(s==n*n)
{
ans=max(ans,sum);
return ;
}
int x=s/n;
int y=s%n;
if(ch[x][y]=='.'&&check(x,y))
{
ch[x][y]='%';
dfs(s+1,sum+1);
ch[x][y]='.';
}
dfs(s+1,sum);
}
int main(int argc, char const *argv[])
{
ios::sync_with_stdio(false);
while(cin>>n&&n)
{
ans=0;
for(int i=0;i<n;i++)
cin>>ch[i];
dfs(0,0);
cout<<ans<<endl;
}
return 0;
}

ZOJ 1002:Fire Net(DFS+回溯)的更多相关文章

  1. ZOJ 1002 Fire Net(dfs)

    嗯... 题目链接:https://zoj.pintia.cn/problem-sets/91827364500/problems/91827364501 这道题是想出来则是一道很简单的dfs: 将一 ...

  2. zoj 1002 Fire Net 碉堡的最大数量【DFS】

    题目链接 题目大意: 假设我们有一个正方形的城市,并且街道是直的.城市的地图是n行n列,每一个单元代表一个街道或者一块墙. 碉堡是一个小城堡,有四个开放的射击口.四个方向是面向北.东.南和西.在每一个 ...

  3. zoj 1002 Fire Net (二分匹配)

    Fire Net Time Limit: 2 Seconds      Memory Limit: 65536 KB Suppose that we have a square city with s ...

  4. [ZOJ 1002] Fire Net (简单地图搜索)

    题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1002 题目大意: 给你一个n*n的地图,地图上的空白部分可以放棋 ...

  5. ZOJ 1002 Fire Net

    题目大意:有一个4*4的城市,其中一些格子有墙(X表示墙),在剩余的区域放置碉堡.子弹不能穿透墙壁.问最多可以放置几个碉堡,保证它们不会相互误伤. 解法:从左上的顶点开始遍历,如果这个点不是墙,做深度 ...

  6. DFS ZOJ 1002/HDOJ 1045 Fire Net

    题目传送门 /* 题意:在一个矩阵里放炮台,满足行列最多只有一个炮台,除非有墙(X)相隔,问最多能放多少个炮台 搜索(DFS):数据小,4 * 4可以用DFS,从(0,0)开始出发,往(n-1,n-1 ...

  7. [ZJU 1002] Fire Net

    ZOJ Problem Set - 1002 Fire Net Time Limit: 2 Seconds      Memory Limit: 65536 KB Suppose that we ha ...

  8. 素数环(dfs+回溯)

    题目描述: 输入正整数n,把整数1,2...n组成一个环,使得相邻两个数和为素数.输出时从整数1开始逆时针排列并且不能重复: 例样输入: 6 例样输出: 1 4 3 2 5 6 1 6 5 2 3 4 ...

  9. NOJ 1074 Hey Judge(DFS回溯)

    Problem 1074: Hey Judge Time Limits:  1000 MS   Memory Limits:  65536 KB 64-bit interger IO format: ...

  10. HDU 1016 Prime Ring Problem(经典DFS+回溯)

    Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Other ...

随机推荐

  1. js图片时间和倒计时

    图片时间原理        原理:使用定时器每秒获取时间,获取时间的时,分,秒,组成一个6位数的字符串,然后用charAt,截取出对应的字符,图片命名和数字相对应就ok拉 倒计时原理        原 ...

  2. zookeeper集群环境搭建(纯zookeeper)

    1.首先在三台机子上放上zookeeper的解压包,解压. 然后的话zookeeper是依赖于jdk的,那么也应该安装jdk,这里不详细说明了. mv zookeeper-3.4.5 zookeepe ...

  3. 无法获取 vmci 驱动程序版本: 句柄无效

    https://jingyan.baidu.com/article/a3a3f811ea5d2a8da2eb8aa1.html 将 vmci0.present = "TURE" 改 ...

  4. IE浏览器兼容的处理方式之一,使用特殊的注释 <!--[if IE]> ....<![endif]-->

    <!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8&quo ...

  5. Problem C: 平面上的点——Point类 (III)

    Description 在数学上,平面直角坐标系上的点用X轴和Y轴上的两个坐标值唯一确定.现在我们封装一个“Point类”来实现平面上的点的操作. 根据“append.cc”,完成Point类的构造方 ...

  6. Linux文件系统命令 lsof

    命令名:lsof 功能:查看某个端口被占用情况,在配置服务器端口的时候,为了避免冲突,可以通过这个命令查看将要被使用的端口是否被占用. 用法:加-i 参数 eg: renjg@renjg-HP-Com ...

  7. Java 容器

    摘录来至https://www.cnblogs.com/LipeiNet/p/5888513.html https://www.cnblogs.com/acm-bingzi/p/javaMap.htm ...

  8. React 新 Context API 在前端状态管理的实践

    本文转载至:今日头条技术博客 众所周知,React的单向数据流模式导致状态只能一级一级的由父组件传递到子组件,在大中型应用中较为繁琐不好管理,通常我们需要使用Redux来帮助我们进行管理,然而随着Re ...

  9. day 53 练习

    1  <!DOCTYPE html><html lang="en"><head> <meta charset="UTF-8&qu ...

  10. kbmMW授权管理解析(The kbmMW Authorization manager explained)

    从kbmMW v.4.40开始,引入了一个新的非常灵活的授权管理器. 它的目的是为开发人员提供为用户定义资源权限的功能,这是一个可选功能,将现有的授权事件驱动方案内置到kbmMW中,使授权开发任务更容 ...