oss locks nes scribe line pac %d input poj

題目鏈接：http://poj.org/problem?id=1128

題目：

Description

Consider the following 5 picture frames placed on an 9 x 8 array.

........ ........ ........ ........ .CCC....

EEEEEE.. ........ ........ ..BBBB.. .C.C....

E....E.. DDDDDD.. ........ ..B..B.. .C.C....

E....E.. D....D.. ........ ..B..B.. .CCC....

 
E....E.. D....D.. ....AAAA ..B..B.. ........

E....E.. D....D.. ....A..A ..BBBB.. ........

E....E.. DDDDDD.. ....A..A ........ ........

E....E.. ........ ....AAAA ........ ........

EEEEEE.. ........ ........ ........ ........

    1        2        3        4        5   

Now place them on top of one another starting with 1 at the bottom and ending up with 5 on top. If any part of a frame covers another it hides that part of the frame below.

Viewing the stack of 5 frames we see the following.

.CCC....

ECBCBB..

DCBCDB..

DCCC.B..

D.B.ABAA

D.BBBB.A

DDDDAD.A

E...AAAA

EEEEEE..

In what order are the frames stacked from bottom to top? The answer is EDABC.

Your problem is to determine the order in which the frames are stacked from bottom to top given a picture of the stacked frames. Here are the rules:

1. The width of the frame is always exactly 1 character and the sides are never shorter than 3 characters.

2. It is possible to see at least one part of each of the four sides of a frame. A corner shows two sides.

3. The frames will be lettered with capital letters, and no two frames will be assigned the same letter.

Input

Each input block contains the height, h (h<=30) on the first line and the width w (w<=30) on the second. A picture of the stacked frames is then given as h strings with w characters each.
Your input may contain multiple blocks of the format described above, without any blank lines in between. All blocks in the input must be processed sequentially.

Output

Write the solution to the standard output. Give the letters of the frames in the order they were stacked from bottom to top. If there are multiple possibilities for an ordering, list all such possibilities in alphabetical order, each one on a separate line. There will always be at least one legal ordering for each input block. List the output for all blocks in the input sequentially, without any blank lines (not even between blocks).

Sample Input

9
8
.CCC....
ECBCBB..
DCBCDB..
DCCC.B..
D.B.ABAA
D.BBBB.A
DDDDAD.A
E...AAAA
EEEEEE..

Sample Output

EDABC

題意：層層覆蓋，從最下面一層開始輸出所有可能的覆蓋順序。

題解：先找到每個圈的每條邊（每個圈都有4條邊），然後判斷並進行構圖，構完圖後就跑DFS，這裏拓撲排序就像個工具，跑DFS的時候註意標記的恢復。

 1 #include <cstdio>
 2 #include <cstring>
 3 using namespace std;
 4 
 5 int n;
 6 const int N=30;
 7 bool E[N][N];
 8 char m[N][N],ans[N];
 9 int vis[N],l[N],r[N],u[N],d[N],in[N],v[N];
10 
11 void init(){
12     n=0;
13     memset(E,0,sizeof(E));
14     for(int i=0;i<30;i++){
15         l[i]=u[i]=30;
16         r[i]=d[i]=0;
17         vis[i]=in[i]=v[i]=0;
18     }
19 }
20 
21 void build(int i,int j,int k){
22     if((k+‘A‘)!=m[i][j]){
23         if(!E[k][m[i][j]-‘A‘]){
24             E[k][m[i][j]-‘A‘]=1;
25             in[m[i][j]-‘A‘]++;
26         }
27     }
28 }
29 
30 void dfs(int st,int cnt){
31     ans[cnt]=st+‘A‘;
32     v[st]=1;
33     if(cnt==n){
34         for(int i=1;i<=n;i++) printf("%c",ans[i]);
35         printf("\n");
36         v[st]=0;
37         return ;
38     }
39     int q[30],c=0;
40     for(int i=0;i<26;i++){
41         if(E[st][i]) in[i]--;
42         if(vis[i]&&!in[i]&&!v[i]) q[c++]=i;
43     }
44     for(int i=0;i<c;i++) dfs(q[i],cnt+1);
45     for(int i=0;i<26;i++) if(E[st][i]) in[i]++;
46     v[st]=0;
47 }
48 
49 int main(){
50     int h,w;
51     while(scanf("%d%d",&h,&w)!=EOF){
52         init();
53         for(int i=1;i<=h;i++) scanf("%s",m[i]+1);
54 
55         for(int i=1;i<=h;i++){
56             for(int j=1;j<=w;j++){
57                 if(m[i][j]!=‘.‘){
58                     int c=m[i][j]-‘A‘;
59                     if(!vis[c]) vis[c]=1,n++;
60                     if(i<u[c]) u[c]=i;
61                     if(i>d[c]) d[c]=i;
62                     if(j>r[c]) r[c]=j;
63                     if(j<l[c]) l[c]=j;
64                 }
65             }
66         }
67         for(int k=0;k<26;k++){
68             int i,j;
69             if(l[k]==30) continue;
70             i=u[k];
71             for(int j=l[k];j<=r[k];j++) build(i,j,k);
72             i=d[k];
73             for(j=l[k];j<=r[k];j++) build(i,j,k);
74             j=l[k];
75             for(i=u[k];i<=d[k];i++) build(i,j,k);
76             j=r[k];
77             for(i=u[k];i<=d[k];i++) build(i,j,k);
78         }
79         for(int i=0;i<26;i++){
80             if(vis[i]&&!in[i]) dfs(i,1);
81         }
82     }
83     return 0;
84 }

POJ 1128 Frame Stacking（拓撲排序+DFS+構圖）