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. jquery过滤器之:contains()、.filter()

    :contains 选择器选取包含指定字符串的元素. 该字符串可以是直接包含在元素中的文本,或者被包含于子元素中. 经常与其他元素/选择器一起使用,来选择指定的组中包含指定文本的元素,如: $(&qu ...

  2. android中的Dialog

    一.Dialog概述 二.使用系统自带的Dialog 1.新建Builder AlertDialog.Builder builder = new AlertDialog.Builder(StoryAc ...

  3. 1021上课演练----SQL注入与避免(银行系统)

    package com.bank; import java.sql.CallableStatement; import java.sql.Connection; import java.sql.Dat ...

  4. ado.net 属性扩展 综合练习

    实现数据查询,添加,删除,修改各项功能 业务逻辑层: using System; using System.Collections.Generic; using System.Linq; using ...

  5. centos6.6 安装 LXC

    LXC,简称Linux containers是docker基础,无奈只能先学习LXC.LXC用途就不多讲,这里只讲LXC的安装以及用途吧! LXC 需要用在内核2.6.27以上 这个可以用 uname ...

  6. Android课程---Android Studio简单设置

    Android Studio 简单设置 界面设置 默认的 Android Studio 为灰色界面,可以选择使用炫酷的黑色界面.Settings-->Appearance-->Theme, ...

  7. Ubuntu下VIM的安装及其配置——Linux篇

    一.Ubuntu系统默认内置: 实际上ubuntu默认没有安装老版本的vi,只装了vim.vi是vim.tiny(vim的最小化版本,不含 GUI,并且仅含有一小部分功能,并且默认与vi兼容.此软件包 ...

  8. linux 解压

    .tar解包:tar xvf FileName.tar打包:tar cvf FileName.tar DirName(注:tar是打包,不是压缩!)-------------------------- ...

  9. Flink - RocksDBStateBackend

    如果要考虑易用性和效率,使用rocksDB来替代普通内存的kv是有必要的 有了rocksdb,可以range查询,可以支持columnfamily,可以各种压缩 但是rocksdb本身是一个库,是跑在 ...

  10. 混合App 框架选型

    个人一直想做个App,但是学习 IOS 的过程发现 原生的做界面还是听麻烦的就放弃了.后来就转到混合式App阵营了 混合式App 定义 Hybrid App(混合模式移动应用)是指介于web-app. ...