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  n*n的地图,'.'表示空地,'X'表示墙,现在往地图上放置炮塔,要求两两炮塔不能同行同列(除非之间有墙),给定一种地图,问这个地图最多可以放置多少个炮塔?
解析 AC代码
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <sstream>
#include <queue>
#include <stack>
#include <vector>
#include <algorithm>
const int maxn=;
using namespace std;
typedef long long ll;
int n;
int ans; //最多可放炮塔数
char map[][];
int canput(int r,int c) //判断(r,c)位置能否放置炮塔
{
int i; //因为行列是递增的,因此只需向上和向左搜索是否有碉堡即可
for(i=r-;i>=&&map[i][c]!='X';i--) //向上
{
if(map[i][c]=='O')
return ;
}
for(i=c-;i>=&&map[r][i]!='X';i--) // 向左
{
if(map[r][i]=='O')
return ;
}
return ;
}
void solve(int k,int current)
{
int x,y;
if(k==n*n) //整个图判断完毕
{
if(current>ans) //更新最优解
{
ans=current;
return;
}
}
else
{
x=k/n; //将单元数转化为坐标
y=k%n;
if(map[x][y]=='.'&& canput(x,y)==) //如果单元格可以放置炮塔
{
map[x][y]='O'; //放置一个炮塔
solve(k+,current+); //递归到下一个单元格
map[x][y]='.'; //恢复单元格 ,回溯
}
solve(k+,current); //本单元格不能放置炮塔
}
}
int main(int argc, char const *argv[])
{
while(scanf("%d",&n)!=EOF&&n)
{
ans=;
for(int i=;i<n;i++)
{
scanf("%s",map[i]);
}
solve(,);
printf("%d\n",ans);
}
return ;
}

ZOJ 1002 DFS的更多相关文章

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. 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 s ...

  5. [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 ...

  6. 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 ...

  7. POJ 1979 Red and Black (zoj 2165) DFS

    传送门: poj:http://poj.org/problem?id=1979 zoj:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problem ...

  8. HDU 1010 Tempter of the Bone (ZOJ 2110) DFS+剪枝

    传送门: HDU:http://acm.hdu.edu.cn/showproblem.php?pid=1010 ZOJ:http://acm.zju.edu.cn/onlinejudge/showPr ...

  9. POJ 1562 Oil Deposits (HDU 1241 ZOJ 1562) DFS

    现在,又可以和她没心没肺的开着玩笑,感觉真好. 思念,是一种后知后觉的痛. 她说,今后做好朋友吧,说这句话的时候都没感觉.. 我想我该恨我自己,肆无忌惮的把她带进我的梦,当成了梦的主角. 梦醒之后总是 ...

随机推荐

  1. C++ 知识点总结复习

    C++ 1.C++是静态类型语言,使用静态类型的编程语言是在编译时执行类型检查,而不是在运行时执行类型检查. 2.面向对象程序设计 C++ 完全支持面向对象的程序设计,包括面向对象开发的四大特性: 封 ...

  2. Java8函数之旅 (八) - 组合式异步编程

    前言 随着多核处理器的出现,如何轻松高效的进行异步编程变得愈发重要,我们看看在java8之前,使用java语言完成异步编程有哪些方案. JAVA8之前的异步编程 继承Thead类,重写run方法 实现 ...

  3. vue监听scroll使用报错的解决办法

    错误说明:在切换路由以后,依旧在其他页面触发了scroll有关的函数, 错误原因:在spa项目中,window对象是不变的,所以每次使用后需要销毁. 解决办法:vue的生命周期destroyed中销毁 ...

  4. 一个好用的PHOTOSHOP切图插件(CutterMan插件下载)

    请关注CutterMan官方微博,分享本站点到自己微博中@Cutterman,私信TA,就有啦~~ 下载地址:http://www.cutterman.cn/ 也许你兴冲冲的下载了,然后发现安装不上, ...

  5. SQLServer 查看SQL语句的执行时间

    在MSSQL Server中通过查看SQL语句执行所用的时间,来衡量SQL语句的性能. 通过设置STATISTICS我们可以查看执行SQL时的系统情况.选项有PROFILE,IO ,TIME.介绍如下 ...

  6. jQuery DOM 元素方法 (十)

    函数 描述 .get() 获得由选择器指定的 DOM 元素. .index() 返回指定元素相对于其他指定元素的 index 位置. .size() 返回被 jQuery 选择器匹配的元素的数量. . ...

  7. 从0到1搭建spark集群---企业集群搭建

    今天分享一篇从0到1搭建Spark集群的步骤,企业中大家亦可以参照次集群搭建自己的Spark集群. 一.下载Spark安装包 可以从官网下载,本集群选择的版本是spark-1.6.0-bin-hado ...

  8. BitCoin p2p通信过程

    众所周知,Bitcoin是建立在p2p网络上的,但是具体的通信过程一直没有搞懂,所以特意去bitcoin的Developer Guid上去了解了一下.由于本人英文水平有限,理解难免有偏差的地方,希望大 ...

  9. mysql之数据操作

    一 介绍 MySQL数据操作: DML 在MySQL管理软件中,可以通过SQL语句中的DML语言来实现数据的操作,包括 使用INSERT实现数据的插入 UPDATE实现数据的更新 使用DELETE实现 ...

  10. ionic2 安装(一)

    1.安装java JDK 2.安装nodejs 3.安装最新版ionic 指令:npm install ionic@latest 4.安装cordova 指令:npm install -g cordo ...