poj3905

Perfect Election

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 962 Accepted: 429

Description

In a country (my memory fails to say which), the candidates {1, 2 ..., N} are running in the parliamentary election. An opinion poll asks the question "For any two candidates of your own choice, which election result would make you happy?". The accepted answers are shown in the table below, where the candidates i and j are not necessarily different, i.e. it may happen that i=j. There are M poll answers, some of which may be similar or identical. The problem is to decide whether there can be an election outcome (It may happen that all candidates fail to be elected, or all are elected, or only a part of them are elected. All these are acceptable election outcomes.) that conforms to all M answers. We say that such an election outcome is perfect. The result of the problem is 1 if a perfect election outcome does exist and 0 otherwise.

Input

Each data set corresponds to an instance of the problem and starts with two integral numbers: 1≤N≤1000 and 1≤M≤1000000. The data set continues with M pairs ±i ±j of signed numbers, 1≤i,j≤N. Each pair encodes a poll answer as follows: 

Accepted answers to the poll questionEncodingI would be happy if at least one from i and j is elected.+i +jI would be happy if at least one from i and j is not elected.-i -jI would be happy if i is elected or j is not elected or both events happen.+i -jI would be happy if i is not elected or j is elected or both events happen.-i +j

The input data are separated by white spaces, terminate with an end of file, and are correct.

Output

For each data set the program prints the result of the encoded election problem. The result, 1 or 0, is printed on the standard output from the beginning of a line. There must be no empty lines on output.

Sample Input

3 3  +1 +2  -1 +2  -1 -3 2 3  -1 +2  -1 -2  +1 -2 2 4  -1 +2  -1 -2  +1 -2  +1 +2 2 8  +1 +2  +2 +1  +1 -2  +1 -2  -2 +1  -1 +1  -2 -2  +1 -1

Sample Output

1101

Hint

For the first data set the result of the problem is 1; there are several perfect election outcomes, e.g. 1 is not elected, 2 is elected, 3 is not elected. The result for the second data set is justified by the perfect election outcome: 1 is not elected, 2 is not elected. The result for the third data set is 0. According to the answers -1 +2 and -1 -2 the candidate 1 must not be elected, whereas the answers +1 -2 and +1 +2 say that candidate 1 must be elected. There is no perfect election outcome. For the fourth data set notice that there are similar or identical poll answers and that some answers mention a single candidate. The result is 1.

题意：有n个候选人，m组要求，每组要求关系到候选人中的两个人，“+i +j”代表i和j中至少有一人被选中，“-i -j”代表i和j中至少有一人不被选中。“+i -j”代表i被选中和j不被选中这两个事件至少发生一个，“-i +j”代表i不被选中和j被选中这两个事件至少发生一个。问是否存在符合所有m项要求的方案存在。

代码:

var
  a:array [1..2010,1..2010] of longint;
  low,dfn,st,c,belong:array [1..2010] of longint;
  v,f:array [1..2010] of boolean;
  k:array [1..2000,1..2000] of boolean;
  i,j,m,n,x,y,d,xx,yy,t,p,tt,c1,c2:longint;
  fff:boolean;


function min(x,y:longint):longint;
begin
  if x>y then
  exit(y);
  exit(x);
end;


function pp:longint;
begin
  f[st[t]]:=false;
  dec(t);
  exit(st[t+1]);
end;


procedure add(x,y:longint);
begin
  inc(c[x]);
  a[x,c[x]]:=y;
end;


procedure check;
var
  i:longint;
begin
  for i:=1 to n do
  if belong[i]=belong[n+i] then
  begin
    fff:=false;
    exit;
  end;
end;


procedure tarjan(x:longint);
var
  i,y:longint;
begin
  inc(d);
  low[x]:=d;
  dfn[x]:=d;
  inc(t);
  st[t]:=x;
  f[x]:=true;
  for i:=1 to c[x] do
  begin
    if not v[a[x,i]] then
    begin
      v[a[x,i]]:=true;
      tarjan(a[x,i]);
      low[x]:=min(low[x],low[a[x,i]]);
    end else
    begin
      if f[a[x,i]] then
      low[x]:=min(low[x],dfn[a[x,i]]);
    end;
  end;
  if dfn[x]=low[x] then
  begin
    inc(p);
    belong[x]:=p;
    y:=x-1;
    while x<>y do
    begin
      y:=pp;
      belong[y]:=p;
    end;
  end;
end;


begin


  while not eof do
  begin
    t:=0;
    p:=0;
    d:=0;
    fillchar(belong,sizeof(belong),0);
    fillchar(low,sizeof(low),0);
    fillchar(dfn,sizeof(dfn),0);
    fillchar(st,sizeof(st),0);
    fillchar(f,sizeof(f),false);
    fillchar(v,sizeof(v),false);
    fillchar(a,sizeof(a),0);
    fillchar(c,sizeof(c),0);
    fillchar(k,sizeof(k),true);
    read(n,m);
    for i:=1 to m do
    begin
      read(x,y);
      if x>0 then
      begin
        if y>0 then
        begin
          add(n+x,y);
          add(n+y,x);
        end else
        begin
          add(n+x,n+abs(y));
          add(abs(y),x);
        end;
      end else
      begin
        if y>0 then
        begin
         add(abs(x),y);
         add(n+y,n+abs(x));
        end
        else
        begin
          add(abs(x),n+abs(y));
          add(abs(y),n+abs(x));
        end;
      end;
    end;
    for i:=1 to n*2 do
    if not v[i] then
    begin
      v[i]:=true;
      tarjan(i);
    end;
    fff:=true;
    check;
    if fff then
    writeln('1') else
    writeln('0');
    readln;
  end;
end.