ZOJ 1002:Fire Net(DFS+回溯)
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
- 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 ...
- 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 ...
- 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 ...
- [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 ...
- 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 ...
- 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 ...
- [ZJU 1002] Fire Net
ZOJ Problem Set - 1002 Fire Net Time Limit: 2 Seconds Memory Limit: 65536 KB Suppose that we ha ...
- 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 ...
- NOJ 1074 Hey Judge(DFS回溯)
Problem 1074: Hey Judge Time Limits: 1000 MS Memory Limits: 65536 KB 64-bit interger IO format: ...
- 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
- Matlab函数
any() 相当于或操作,只要有1,就返回1 all() 相当于与操作,只要有0,就返回0 C = union(A,B): C为A和B的并集.去掉相同元素. C = intersect(A,B) C为 ...
- linux下/proc/diskstats文件详解
每一列的含义分别为: 第一列为 设备号 (number of issued reads. This is the total number of reads completed successfull ...
- 基于Quartz.NET 实现可中断的任务(转)
Quartz.NET 是一个开源的作业调度框架,非常适合在平时的工作中,定时轮询数据库同步,定时邮件通知,定时处理数据等. Quartz.NET 允许开发人员根据时间间隔(或天)来调度作业.它实现了作 ...
- ASP.Net MVC(3) 之Razor视图引擎的基础语法
Razor视图引擎的基础语法: 1.“_”开头的cshtml文档将不能在服务器上访问,和asp.net中的config文档差不多. 2.Razor语法以@开头,以@{}进行包裹. 3.语法使用: 注释 ...
- tomcat vue webpack vue-router 404
社区已经有结局方案了, http://blog.csdn.net/hayre/article/details/70145513
- poj3261
题解: 同bzoj1717 代码: #include<bits/stdc++.h> using namespace std; ,P2=,P=; int a1[P],num[P],a2[P] ...
- [Codeforces797F]Mice and Holes
Problem n个老鼠,m个洞,告诉你他们的一维坐标和m个洞的容量限制,问最小总距离. Solution 用dp[i][j]表示前i个洞,进了前j个老鼠的最小代价 dp[i][j]=min(dp[i ...
- Instruments leak黑魔法定位内存泄漏
leak是一款很赞的内存检查的工具,但在使用的过程中有点繁琐,至少有些底层的泄漏笔者还是不知道如何下手 下面介绍一下简单leak的使用: 首先你要确认你的target不会被拒绝,确保profile是d ...
- bootstrap动态生成层级ul-li 新闻预览 常用方法
<div class="row" id="add-withinfosortId-row" style="display: none"& ...
- DevExpress WPF v18.2新版亮点(四)
行业领先的.NET界面控件2018年第二次重大更新——DevExpress v18.2日前正式发布,本站将以连载的形式为大家介绍新版本新功能.本文将介绍了DevExpress WPF v18.2的新功 ...