HDU 4770 Lights Against DudelyLights
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
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.
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
If there are no vulnerable rooms, print 0.
If Dumbledore has no way to light up all vulnerable rooms, print -1.
##
##
2 3
#..
..#
3 3
###
#.#
###
0 0
2
-1
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
- hdu 4770 Lights Against Dudely(回溯)
pid=4770" target="_blank" style="">题目链接:hdu 4770 Lights Against Dudely 题 ...
- 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 ...
- HDU 4770 Lights Against Dudely
Lights Against Dudely Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Ot ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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
- 【android design】android常用设计资源
一.概述 大部分程序员擅长开发,但是对于设计却知之甚少.这直接导致,程序员在初期开发出来的应用(大多为兴趣或实用导向)中看不中用.因此,有必要搜集整合一些设计资源,这样既能减轻程序员在设计上所耗费的时 ...
- [LintCode] Coins in a Line II 一条线上的硬币之二
There are n coins with different value in a line. Two players take turns to take one or two coins fr ...
- "我爱记单词"测试报告兼功能展示
"我爱记单词"测试报告兼功能展示 前言: 我们大部分的测试都是一边开发一边完成的,这里给出软件开发基本完成后在使用时的一些测试例子. 一.背景介绍 我们的数据库中一共有10个表: ...
- 类 class
<?php class Person{ //定义一个类 public $name; //定义属性 public $age; function say(){ //定义办法 echo "m ...
- jQuery上定义插件并重设插件构造函数
插件alert的全部代码,每个插件都定义在如下类似的作用域中: +function ($) { 'use strict'; // ALERT CLASS DEFINITION // ========= ...
- php数字操作,高精度函数,保留计算结果小数位
$l = 45456.51; $r = 455778.44; $e = '100.00'; $f= '500.00'; $res = bcadd($l, $r,3);//小数点后的位数,精度就是由这个 ...
- XE5 ImageList的BUG?
今天做界面, 在imagelist里加载一个带有半透明通道的PNG图, 结果发现图片居然发暗, 如下: 原图: IDE里加载以后的图: 明显变暗...查询了源码, 无果 然后又用2010去测试, 发现 ...
- http://www.cnblogs.com/softidea/p/5631763.html
http://www.cnblogs.com/softidea/p/5631763.html
- AFNetWorking 队列请求
我们在开发过程中,经常会遇到有些页面不止一个网络请求,有时候需要两个三个甚至更多,这个时候我们就需要队列请求,下边是GET请求的多个请求放在队列里边: NSURL *url = [NSURL URLW ...
- free-library-converts-2d-image-to-3d
http://www.i-programmer.info/news/105-artificial-intelligence/4917-free-library-converts-2d-image-to ...