2102: [Usaco2010 Dec]The Trough Game

2102: [Usaco2010 Dec]The Trough Game

Time Limit: 10 Sec  Memory Limit: 64 MB
Submit: 117  Solved: 84
[Submit][Status]

Description

Farmer John and Bessie are playing games again. This one has to do with troughs of water. Farmer John has hidden N (1 <= N <= 20) troughs behind the barn, and has filled some of them with food. Bessie has asked M (1 <= M <= 100) questions of the form, "How many troughs from this list (which she recites) are filled?". Bessie needs your help to deduce which troughs are actually filled. Consider an example with four troughs where Bessie has asked these questions (and received the indicated answers): 1) "How many of these troughs are filled: trough 1" --> 1 trough is filled 2) "How many of these troughs are filled: troughs 2 and 3" --> 1 trough is filled 3) "How many of these troughs are filled: troughs 1 and 4" --> 1 trough is filled 4) "How many of these troughs are filled: troughs 3 and 4" --> 1 trough is filled From question 1, we know trough 1 is filled. From question 3, we then know trough 4 is empty. From question 4, we then know that trough 3 is filled. From question 2, we then know that trough 2 is empty. 求N位二进制数X，使得给定的M个数，满足X and Bi=Ci ,Bi ci分别是读入的两个数

Input

* Line 1: Two space-separated integers: N and M * Lines 2..M+1: A subset of troughs, specified as a sequence of contiguous N 0's and 1's, followed by a single integer that is the number of troughs in the specified subset that are filled.

Output

* Line 1: A single line with: * The string "IMPOSSIBLE" if there is no possible set of filled troughs compatible with Farmer John's answers. * The string "NOT UNIQUE" if Bessie cannot determine from the given data exactly what troughs are filled. * Otherwise, a sequence of contiguous N 0's and 1's specifying which troughs are filled.

Sample Input

4 4
1000 1
0110 1
1001 1
0011 1

Sample Output

1010

HINT

 

Source

Silver

 

题解：一上来居然没有别的想法——只有暴力。。。然后写了个纯粹的二进制穷举，然后，然后，然后，居然AC了？！？！44ms也是醉大了= =

type
    point=^node;
    node=record
               g:longint;
               next:point;
    end;
var
   i,j,k,l,m,n,t:longint;
   a:array[0..10000] of point;
   b,c,d:array[0..10000] of longint;
   c1,c2:char;
procedure add(x,y:longint);inline;
          var p:point;
          begin
               new(p);p^.g:=y;
               p^.next:=a[x];a[x]:=p;
          end;
procedure dfs(x:longint);inline;
          var i,j,k,l:longint;p:point;
          begin
               if x>n then
                  begin
                       for i:=1 to m do if b[i]>0 then exit;
                       if t=0 then
                          begin
                               for i:=1 to n do d[i]:=c[i];
                               t:=1;
                          end
                       else
                           begin
                                writeln('NOT UNIQUE');
                                halt;
                           end;
                  end
               else
                   begin
                        p:=a[x];l:=0;
                        while p<>nil do
                              begin
                                   if b[p^.g]=0 then
                                      begin
                                           l:=1;
                                           break;
                                      end;
                                   p:=p^.next;
                              end;
                        if l=0 then
                           begin
                                p:=a[x];
                                while p<>nil do
                                      begin
                                           dec(b[p^.g]);
                                           p:=p^.next;
                                      end;
                                c[x]:=1;
                                dfs(x+1);
                                p:=a[x];
                                while p<>nil do
                                      begin
                                           inc(b[p^.g]);
                                           p:=p^.next;
                                      end;
                           end;
                        c[x]:=0;
                        dfs(x+1);
                   end;
          end;

begin
     readln(n,m);
     for i:=1 to m do a[i]:=nil;
     for i:=1 to m do
         begin
              for j:=1 to n do
                  begin
                       read(c1);
                       if c1='1' then add(j,i);
                  end;
              readln(b[i]);
         end;
     t:=0;
     dfs(1);
     IF t=0 then write('IMPOSSIBLE') else for i:=1 to n do write(d[i]);
     writeln;
     readln;
end.