HDU 4770 Lights Against Dudely(枚举所有状态 当然壮压dp会很简单)

Lights Against Dudely

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 450    Accepted Submission(s): 134


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

 

题目大意：点表示可以照亮的灯，普通的一盏灯可以照亮自己的位置和自己上和右的位置。特殊的灯可以照亮的位置可以将普通灯旋转。问你最少使用多少灯照亮所有的点。
注意：
1.每次最多有一个特殊灯。
2.总共最多是15个点。
3.一个点如果被一盏灯占住了，则不可以放灯，但是被照亮了，可以继续放灯。

解题思路：当时开始讨论的思路是暴利，时间复杂度，O（2^14*15*4）大概是10^6，然后可以加一个剪枝，top*3<h可以直接跳出。但最后是用壮压dp写出来的，直接0ms一A

题目地址：Lights Against Dudely

下面是自己的挫代码。。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n,m,h;    //h是点的个数
char map1[205][205];  //地图
int dian[20][2];   //点的坐标
int dir[4][2]= {{-1,0},{0,1},{1,0},{0,-1}}; //四个方向
int vis[20];
int sav[20];
int visi[25];  //是否全部点都被照亮
int map2[205][205];   //存放每个点的顺序
int north[25],south[25],west[25],east[25];   //记录每个点的上下左右点的在点里的位置

int ok1(int p)   //只能右上
{
int x1,y1,x2,y2,x,y;
x=dian[p][0],y=dian[p][1];
x1=x+dir[0][0],y1=y+dir[0][1];
x2=x+dir[1][0],y2=y+dir[1][1];
if(map1[x1][y1]=='.'&&map1[x2][y2]=='.')
return 1;
return 0;
}

int ok2(int p,int i)   //特殊点
{
int x1,y1,x2,y2,x,y,j;
x=dian[p][0],y=dian[p][1];
j=(i+1)%4;
x1=x+dir[i][0],y1=y+dir[i][1];
x2=x+dir[j][0],y2=y+dir[j][1];
if(map1[x1][y1]=='.'&&map1[x2][y2]=='.') return 1;
return 0;
}

int cover()
{
int i;
for(i=0; i<h; i++)
if(!visi[i])
return 0;
return 1;
}

int solve()
{
int i,j,k;
int top=0;
for(i=0; i<h; i++)
if(vis[i])
sav[top++]=i;
if(top*3<h) return 20;  //不可能照亮
for(k=0; k<=3; k++)
{
int flag;
for(i=0; i<top; i++)
{
if(!ok2(sav[i],k)) continue;
int ii=sav[i];
memset(visi,0,sizeof(visi));
visi[ii]=1;
if(k==0) visi[north[ii]]=1,visi[east[ii]]=1;
else if(k==1) visi[south[ii]]=1,visi[east[ii]]=1;
else if(k==2) visi[south[ii]]=1,visi[west[ii]]=1;
else visi[north[ii]]=1,visi[west[ii]]=1;

flag=0;
for(j=0; j<top; j++)
{
if(i==j) continue;
if(!ok1(sav[j]))
{
flag=1;
break;
}
int jj=sav[j];
visi[jj]=1;
visi[north[jj]]=1,visi[east[jj]]=1;
}

if(flag) continue;
if(cover()) return top;
}
}
return 20;
}

int main()
{
int i,j;
while(scanf("%d%d",&n,&m))
{
if(n==0&&m==0) break;
for(i=1; i<=n; i++)
scanf("%s",map1[i]+1);

h=0;   //点的个数
for(i=1; i<=n; i++)
for(j=1; j<=m; j++)
{
if(map1[i][j]=='.')
{
map2[i][j]=h;
dian[h][0]=i;
dian[h++][1]=j;
}
else map2[i][j]=20;
}

for(i=0;i<h;i++)
north[i]=south[i]=west[i]=east[i]=20;
for(i=0; i<=m+1; i++)
{
map1[0][i]='.',map1[n+1][i]='.';
map2[0][i]=20,map2[n+1][i]=20;
}
for(i=0; i<=n+1; i++)
{
map1[i][0]='.',map1[i][m+1]='.';
map2[i][0]=20,map2[i][m+1]=20;
}

for(i=1; i<=n; i++)
for(j=1; j<=m; j++)
{
if(map1[i][j]=='.')
{
if(map1[i][j-1]=='.')
{
west[map2[i][j]]=map2[i][j-1];
east[map2[i][j-1]]=map2[i][j];
}
if(map1[i][j+1]=='.')
{
east[map2[i][j]]=map2[i][j+1];
west[map2[i][j+1]]=map2[i][j];
}
if(map1[i-1][j]=='.')
{
north[map2[i][j]]=map2[i-1][j];
south[map2[i-1][j]]=map2[i][j];
}
if(map1[i+1][j]=='.')
{
south[map2[i][j]]=map2[i+1][j];
north[map2[i+1][j]]=map2[i][j];
}
}
}

int s=(1<<h)-1;
int res=20;
if(s==0)
{
puts("0");
continue;
}

for(i=1; i<=s; i++)
{
int tmp=i;
for(j=0; j<h; j++)
{
vis[j]=tmp&1;
tmp>>=1;
}

res=min(res,solve());
}
if(res==20) { puts("-1"); continue;}
else   cout<<res<<endl;
}
return 0;
}

/*
2 2
##
##
2 3
#..
..#
3 3
###
#.#
###
4 4
.##.
####
####
.##.
4 4
.##.
..#.
..#.
..#.
5 4
#.##
..#.
#.#.
#..#
....
2 4
.##.
####
2 4
.###
####
0
2
-1
-1
5
-1
2
1
*/

//93MS