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。

Lights Against Dudely

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1360    Accepted Submission(s): 392

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

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思路：暴力枚举每一个款式特殊灯的位置（转自：http://blog.csdn.net/u014737310/article/details/39249581）

好像状压更简单：

http://www.cnblogs.com/kuangbin/p/3416163.html

  1 #include<iostream>
  2 #include<cstring>
  3 #include<cstdlib>
  4 #include<cstdio>
  5 #include<algorithm>
  6 #include<cmath>
  7 #include<queue>
  8 #include<map>
  9 #include<string>
 10 //#include<pair>
 11
 12 #define N 205
 13 #define M 15
 14 #define mod 10000007
 15 //#define p 10000007
 16 #define mod2 100000000
 17 #define ll long long
 18 #define LL long long
 19 #define maxi(a,b) (a)>(b)? (a) : (b)
 20 #define mini(a,b) (a)<(b)? (a) : (b)
 21
 22 using namespace std;
 23
 24 int n,m;
 25 int ans;
 26 int k;
 27 int vis[N][N];
 28 char s[N][N];
 29 int c[N];
 30 int step[4][3][2]={  {-1,0,0,0,0,1},
 31                     {0,-1,-1,0,0,0},
 32                     {0,-1,0,0,1,0},
 33                     {0,0,0,1,1,0}
 34                  };
 35 typedef struct
 36 {
 37     int x;
 38     int y;
 39 }PP;
 40
 41 PP p[N];
 42
 43 void change(int x,int y,int st[3][2],int s)
 44 {
 45     int i;
 46     int nx,ny;
 47     for(i=0;i<3;i++){
 48         nx=x+st[i][0];
 49         ny=y+st[i][1];
 50         if(nx<0 || nx>=n) continue;
 51         if(ny<0 || ny>=m) continue;
 52
 53         vis[nx][ny]=s;
 54     }
 55 }
 56
 57 int judge(int x,int y,int st[3][2])
 58 {
 59     int i;
 60     int nx,ny;
 61     for(i=0;i<3;i++){
 62         nx=x+st[i][0];
 63         ny=y+st[i][1];
 64         if(nx<0 || nx>=n || ny<0 || ny>=m) continue;
 65         if(s[nx][ny]==‘#‘){
 66             return 0;
 67         }
 68     }
 69     return 1;
 70 }
 71
 72 void ini()
 73 {
 74     int i,j;
 75     memset(c,0,sizeof(c));
 76     ans=20;
 77     k=0;
 78     for(i=0;i<n;i++){
 79         scanf("%s",s[i]);
 80     }
 81     for(i=n-1;i>=0;i--){
 82         for(j=0;j<m;j++){
 83             if(s[i][j]==‘.‘){
 84                 k++;
 85                 p[k].x=i;p[k].y=j;
 86             }
 87         }
 88     }
 89
 90     for(i=1;i<=k;i++){
 91         if(judge(p[i].x,p[i].y,step[0])==1){
 92             c[i]=1;
 93         }
 94         else{
 95             c[i]=-1;
 96         }
 97     }
 98 }
 99
100
101 void dfs(int i,int f,int re)
102 {
103     if(re>=ans){
104         return;
105     }
106     if(i==k+1){
107         ans=re;
108         return;
109     }
110     if(vis[ p[i].x ][ p[i].y ]==1)
111     {
112         dfs(i+1,f,re);
113     }
114     if(f!=i && c[i]==1){
115         change(p[i].x,p[i].y,step[0],1);
116         dfs(i+1,f,re+1);
117         change(p[i].x,p[i].y,step[0],0);
118     }
119 }
120
121 void solve()
122 {
123     int i,j;
124     if(k==0){
125         ans=0;return;
126     }
127     memset(vis,0,sizeof(vis));
128     dfs(1,0,0);
129     for(j=1;j<=3;j++)
130     {
131         memset(vis,0,sizeof(vis));
132         for(i=1;i<=k;i++){
133             if(judge( p[i].x,p[i].y,step[j] )==1){
134                 change(p[i].x,p[i].y,step[j],1);
135                 dfs(1,i,1);
136                 change(p[i].x,p[i].y,step[j],0);
137             }
138         }
139     }
140 }
141
142
143 void out()
144 {
145     if(ans==20) ans=-1;
146     printf("%d\n",ans);
147 }
148
149 int main()
150 {
151    // freopen("data.in","r",stdin);
152     //freopen("data.out","w",stdout);
153     //scanf("%d",&T);
154    // for(int ccnt=1;ccnt<=T;ccnt++)
155    // while(T--)
156     while(scanf("%d%d",&n,&m)!=EOF)
157     {
158         if(n==0 && m==0 ) break;
159         //printf("Case %d: ",ccnt);
160         ini();
161         solve();
162         out();
163     }
164
165     return 0;
166 }