Lights Against Dudely

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 525    Accepted Submission(s): 157

Problem Description
Harry: "But Hagrid. How am I going to pay for all of this? I haven't any money."

Hagrid: "Well there's your money, Harry! Gringotts, the wizard bank! Ain't no safer place. Not one. Except perhaps Hogwarts."

— Rubeus Hagrid to Harry Potter.

  Gringotts Wizarding Bank is the only bank of the wizarding world, and is owned and operated by goblins. It was created by a goblin called Gringott. Its main offices are located in the North Side of Diagon Alley in London, England. In addition to storing money and valuables for wizards and witches, one can go there to exchange Muggle money for wizarding money. The currency exchanged by Muggles is later returned to circulation in the Muggle world by goblins. According to Rubeus Hagrid, other than Hogwarts School of Witchcraft and Wizardry, Gringotts is the safest place in the wizarding world.

  The text above is quoted from Harry Potter Wiki. But now Gringotts Wizarding Bank is not safe anymore. The stupid Dudley, Harry Potter's cousin, just robbed the bank. Of course, uncle Vernon, the drill seller, is behind the curtain because he has the most advanced drills in the world. Dudley drove an invisible and soundless drilling machine into the bank, and stole all Harry Potter's wizarding money and Muggle money. Dumbledore couldn't stand with it. He ordered to put some magic lights in the bank rooms to detect Dudley's drilling machine. The bank can be considered as a N × M grid consisting of N × M rooms. Each room has a coordinate. The coordinates of the upper-left room is (1,1) , the down-right room is (N,M) and the room below the upper-left room is (2,1)..... A 3×4 bank grid is shown below:

Some rooms are indestructible and some rooms are vulnerable. Dudely's machine can only pass the vulnerable rooms. So lights must be put to light up all vulnerable rooms. There are at most fifteen vulnerable rooms in the bank. You can at most put one light in one room. The light of the lights can penetrate the walls. If you put a light in room (x,y), it lights up three rooms: room (x,y), room (x-1,y) and room (x,y+1). Dumbledore has only one special light whose lighting direction can be turned by 0 degree,90 degrees, 180 degrees or 270 degrees. For example, if the special light is put in room (x,y) and its lighting direction is turned by 90 degrees, it will light up room (x,y), room (x,y+1 ) and room (x+1,y). Now please help Dumbledore to figure out at least how many lights he has to use to light up all vulnerable rooms.

Please pay attention that you can't light up any indestructible rooms, because the goblins there hate light.

 
Input
There are several test cases.

In each test case:

The first line are two integers N and M, meaning that the bank is a N × M grid(0<N,M <= 200).

Then a N×M matrix follows. Each element is a letter standing for a room. '#' means a indestructible room, and '.' means a vulnerable room.

The input ends with N = 0 and M = 0

 
Output
For each test case, print the minimum number of lights which Dumbledore needs to put.

If there are no vulnerable rooms, print 0.

If Dumbledore has no way to light up all vulnerable rooms, print -1.
 
Sample Input
2 2
##
##
2 3
#..
..#
3 3
###
#.#
###
0 0
 
Sample Output
0
2
-1
 
Source
2013 Asia Hangzhou Regional Contest


解题思路:杭州的A题,看到网上很多大牛用状态压缩来写,表示YM,虽然我也想到了压缩,不过这一题不用压缩就能写,所以我没用压缩,直接爆搜加优化就可以了。详见代码

#include<iostream>
#include<cstdio>
#include<cstring>
#define N 205
using namespace std;
char maze[N][N];
int x[20],y[20],visit[N][N],ans;
bool vis[N][N];
void recover(int k)
{
int i;
for(i=0;i<k;i++)
visit[x[i]][y[i]]=0,vis[x[i]][y[i]]=false;
}
void dfs(int cnt,int now,int k,int dep)
{
if(cnt>=ans-1&&now!=k||cnt>dep)
//优化3:已经放的灯数比现在的答案大就直接返回
return ;
if(now==k)
{
ans=min(ans,cnt);
return ;
}
for(int i=0;i<k;i++)
{
int flag1=0,flag2=0,flag3=0,flag=0;
if(maze[x[i]-1][y[i]]!='#'&&maze[x[i]][y[i]+1]!='#'&&!vis[x[i]][y[i]]&&(visit[x[i]][y[i]]==0||(visit[x[i]][y[i]+1]==0&&maze[x[i]][y[i]+1]=='.')||(visit[x[i]-1][y[i]]==0&&maze[x[i]-1][y[i]]=='.')))
//vis数组记录该位置放灯了没,visit数组记录该位置是否被照亮,优化2:放下一个灯的时候至少能多照亮一个位置才放灯
{
if(visit[x[i]][y[i]]==0)
flag1=1;
if(visit[x[i]][y[i]+1]==0)
flag2=1;
if(visit[x[i]-1][y[i]]==0)
flag3=1;
flag=flag1+flag2+flag3;
visit[x[i]][y[i]]++;
visit[x[i]-1][y[i]]++;
visit[x[i]][y[i]+1]++;
vis[x[i]][y[i]]=true;
if(maze[x[i]-1][y[i]]=='.'&&maze[x[i]][y[i]+1]=='.')
dfs(cnt+1,now+flag,k,dep);
else if(maze[x[i]-1][y[i]]=='.')
dfs(cnt+1,flag1+flag3+now,k,dep);
else if(maze[x[i]][y[i]+1]=='.')
dfs(cnt+1,flag1+flag2+now,k,dep);
else
dfs(cnt+1,flag1+now,k,dep);
visit[x[i]][y[i]]--;
visit[x[i]-1][y[i]]--;
visit[x[i]][y[i]+1]--;
vis[x[i]][y[i]]=false;
}
}
return ;
}
int main()
{
int m,n,i,j,k,t;
bool flag;
while(scanf("%d%d",&m,&n),m+n)
{
flag=true;
k=t=0;
ans=100;
getchar();
for(i=0;i<=m+1;i++)
for(j=0;j<=n+1;j++)
{
maze[i][j]='@';
visit[i][j]=0;
vis[i][j]=false;
}
for(i=1;i<=m;i++)
{
for(j=1;j<=n;j++)
{
scanf("%c",&maze[i][j]);
if(maze[i][j]=='.')
x[k]=i,y[k++]=j;
}
getchar();
}
if(k==0)
{
printf("0\n");
continue;
}
int Min=k%3?k/3+1:k/3;//优化1:n个可以放灯的点至少需要[n/3]展灯
for(i=0;i<k;i++)
{
if(maze[x[i]-1][y[i]]!='#'&&maze[x[i]][y[i]+1]!='#')
{
visit[x[i]][y[i]]++;
visit[x[i]-1][y[i]]++;
visit[x[i]][y[i]+1]++;
vis[x[i]][y[i]]=true;
if(maze[x[i]-1][y[i]]=='@'&&maze[x[i]][y[i]+1]=='@')
{
for(j=Min;j<=min(ans,k);j++)
dfs(1,1,k,j);
}
else if(maze[x[i]-1][y[i]]=='@'||maze[x[i]][y[i]+1]=='@')
{
for(j=Min;j<=min(ans,k);j++)
dfs(1,2,k,j);
}
else
{
for(j=Min;j<=min(ans,k);j++)
dfs(1,3,k,j);
}
}
recover(k);
flag=true;
if(maze[x[i]][y[i]+1]!='#'&&maze[x[i]+1][y[i]]!='#')
{
visit[x[i]][y[i]]++;
visit[x[i]+1][y[i]]++;
visit[x[i]][y[i]+1]++;
vis[x[i]][y[i]]=true;
if(maze[x[i]+1][y[i]]=='@'&&maze[x[i]][y[i]+1]=='@')
{
for(j=Min;j<=min(ans,k);j++)
dfs(1,1,k,j);
}
else if(maze[x[i]+1][y[i]]=='@'||maze[x[i]][y[i]+1]=='@')
{
for(j=Min;j<=min(ans,k);j++)
dfs(1,2,k,j);
}
else
{
for(j=Min;j<=min(ans,k);j++)
dfs(1,3,k,j);
}
}
recover(k);
flag=true;
if(maze[x[i]+1][y[i]]!='#'&&maze[x[i]][y[i]-1]!='#')
{
visit[x[i]][y[i]]++;
visit[x[i]+1][y[i]]++;
visit[x[i]][y[i]-1]++;
vis[x[i]][y[i]]=true;
if(maze[x[i]+1][y[i]]=='@'&&maze[x[i]][y[i]-1]=='@')
{
for(j=Min;j<=min(ans,k);j++)
dfs(1,1,k,j);
}
else if(maze[x[i]+1][y[i]]=='@'||maze[x[i]][y[i]-1]=='@')
{
for(j=Min;j<=min(ans,k);j++)
dfs(1,2,k,j);
}
else
{
for(j=Min;j<=min(ans,k);j++)
dfs(1,3,k,j);
} }
recover(k);
flag=true;
if(maze[x[i]][y[i]-1]!='#'&&maze[x[i]-1][y[i]]!='#')
{
visit[x[i]][y[i]]++;
visit[x[i]-1][y[i]]++;
visit[x[i]][y[i]-1]++;
vis[x[i]][y[i]]=true;
if(maze[x[i]-1][y[i]]=='@'&&maze[x[i]][y[i]-1]=='@')
{
for(j=Min;j<=min(ans,k);j++)
dfs(1,1,k,j);
}
else if(maze[x[i]-1][y[i]]=='@'||maze[x[i]][y[i]-1]=='@')
{
for(j=Min;j<=min(ans,k);j++)
dfs(1,2,k,j);
}
else
{
for(j=Min;j<=min(ans,k);j++)
dfs(1,3,k,j);
}
}
recover(k);
flag=true;
}
if(ans==100)
printf("-1\n");
else
printf("%d\n",ans);
}
return 0;
}

HDU 4770 Lights Against DudelyLights的更多相关文章

  1. hdu 4770 Lights Against Dudely(回溯)

    pid=4770" target="_blank" style="">题目链接:hdu 4770 Lights Against Dudely 题 ...

  2. HDU 4770 Lights Against Dudely 暴力枚举+dfs

    又一发吐血ac,,,再次明白了用函数(代码重用)和思路清晰的重要性. 11779687 2014-10-02 20:57:53 Accepted 4770 0MS 496K 2976 B G++ cz ...

  3. HDU 4770 Lights Against Dudely

    Lights Against Dudely Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Ot ...

  4. HDU 4770 Lights Against Dudely (2013杭州赛区1001题,暴力枚举)

    Lights Against Dudely Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Ot ...

  5. HDU 4770 Lights Against Dudely(暴力+状压)

    思路: 这个题完全就是暴力的,就是代码长了一点. 用到了状压,因为之前不知道状压是个东西,大佬们天天说,可是我又没学过,所以对状压有一点阴影,不过这题中的状压还是蛮简单的. 枚举所有情况,取开灯数最少 ...

  6. hdu 4770 13 杭州 现场 A - Lights Against Dudely 暴力 bfs 状态压缩DP 难度:1

    Description Harry: "But Hagrid. How am I going to pay for all of this? I haven't any money.&quo ...

  7. hdu 4770(枚举 + dfs爆搜)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4770 思路:由于最多只有15个".",可以直接枚举放置的位置,然后判断是否能够全部 ...

  8. HDOJ 4770 Lights Against Dudely

    状压+暴力搜索 Lights Against Dudely Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K ...

  9. HDU 5820 Lights(扫描线+zkw线段树)

    [题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=5820 [题目大意] 在一个大小为50000*50000的矩形中,有n个路灯. 询问是否每一对路灯之 ...

随机推荐

  1. [CareerCup] 17.13 BiNode 双向节点

    17.13 Consider a simple node-like data structure called BiNode, which has pointers to two other node ...

  2. LeetCode(43. Multiply Strings)

    题目: Given two numbers represented as strings, return multiplication of the numbers as a string. Note ...

  3. Centos 安装了 Wkhtmltopdf 却依旧显示 无法打印pdf

    Odoo里判断wkhtmlpdf是否安装的代码在 openerp/tools/misc.py 文件中: def find_in_path(name): path = os.environ.get('P ...

  4. ckeditor简单的演示

    先把ckeditor文件添加到项目中 然后在页面上引用 <!DOCTYPE html> <html xmlns="http://www.w3.org/1999/xhtml& ...

  5. vbox下Oracle Enterprise liunx5.4虚拟机安装10G RAC实验(四)

    接第3篇 http://www.cnblogs.com/myrunning/p/4003527.html 5.安装配置数据库 5.1安装数据库软件 5.2配置监听 5.3创建ASM磁盘 5.4创建服务 ...

  6. 支付宝APP支付之Java后台生成签名具体步骤

    /** *支付宝支付 * @param orderId 订单编号 * @param actualPay 实际支付金额 * @return */ private String getOrderInfoB ...

  7. 通过Queue的构造函数的可选参数maxsize来设定队列长度

    创建一个"队列"对象 import Queuemyqueue = Queue.Queue(maxsize = 10) Queue.Queue类即是一个队列的同步实现.队列长度可为无 ...

  8. JavaAnnotation和反射简化开发

    Part I 我并不是在卖弄自己的英语有多少的了不起,只不过对Annotation这一次的解释真的很懊恼,“注解”和“注释”这两个对Annotation的翻译我听着不爽,所以全文都用Annotatio ...

  9. venus java高并发框架

    http://www.iteye.com/topic/1118484 因为有 netty.mima等远程框架.包括spring jboss等remoting框架 和阿里的dubbo相比, 没有亮点.. ...

  10. mybatis底层源码分析之--配置文件读取和解析

    现在企业级开发中ssm是很常见的技术标配,mybatis比hibernate轻量了很多,而且学习成本相对较低,简单易上手. 那么,问题来了,简单好用的mybatis底层到底是如何实现的呢?都使用了什么 ...