POJ 3207 Ikki's Story IV

hzwer大神推荐的非常赞的介绍：http://blog.csdn.net/jarjingx/article/details/8521690

简单的看了2-sat……似乎还是挺神奇的东西……等大致刷完几道题再来写总结吧！

而这道题……算是2-sat的超级入门题了吧

不过题目大意也是醉了：圆上顺序排列n个点，现要在一些点间连边，规定边只能在圆内或圆外，求有没有可能不相交 。一开始想的是嗷嗷嗷，圆上两个点的连线怎么可能有什么在圆外在圆内之分，不都是弦么？后来才知道……原来两个点的连线不是直线可以使曲线……

判定是否相交很简单吧……看成是一条直线上四个点，那么如果e1.a<e2.a<e1.b<e2.b就相交啦……

都说是热身题没多难……

type
  arr1=record
    a,b:longint;
  end;
  arr2=record
    toward,next:longint;
  end;

var
  e:array[0..3000]of arr1;
  edge:array[0..3000]of arr2;
  first,dfn,low,scc,p,num1,num2:array[0..3000]of longint;
  chose:array[0..3000]of boolean;
  time,total,tot,n,m,tail,sum:longint;

procedure swap(var x,y:longint);
var
  i:longint;
begin
  i:=x;
  x:=y;
  y:=i;
end;

procedure addedge(j,k:longint);
begin
  inc(total);
  edge[total].toward:=k;
  edge[total].next:=first[j];
  first[j]:=total;
end;

procedure add(j,k:longint);
begin
  addedge(j,k);
  addedge(k,j);
end;

function check(x,y:arr1):boolean;
begin
  if (x.a<y.a) and (y.a<x.b) and (x.b<y.b) then exit(true);
  exit(false);
end;

procedure tarjan(x:longint);
var
  i,too,j:longint;
begin
  inc(time);
  dfn[x]:=time;
  low[x]:=time;
  chose[x]:=true;
  inc(tail);
  p[tail]:=x;
  i:=first[x];
  while i>0 do begin
    too:=edge[i].toward;
    if dfn[too]=0 then begin
      tarjan(too);
      if low[too]<low[x] then low[x]:=low[too];
    end
    else
      if dfn[too]<low[x] then low[x]:=dfn[too];
    i:=edge[i].next;
  end;
  if dfn[x]=low[x] then begin
    inc(sum);
    repeat
      j:=p[tail];
      dec(tail);
      chose[j]:=false;
      scc[j]:=sum;
    until j=x;
  end;
end;

procedure into;
var
  i,j:longint;
begin
  readln(n,m);
  for i:=1 to m do begin
    read(e[i].a,e[i].b);
    if e[i].a>e[i].b then swap(e[i].a,e[i].b);
    inc(tot);
    num1[i]:=tot;
    inc(tot);
    num2[i]:=tot;
  end;
  for i:=1 to m-1 do
    for j:=i+1 to m do
      if check(e[i],e[j]) or check(e[j],e[i]) then begin
        add(num1[i],num2[j]);
        add(num2[i],num1[j]);
      end;
end;

function work:boolean;
var
  i:longint;
begin
  for i:=1 to tot do
    if dfn[i]=0 then tarjan(i);
  for i:=1 to m do
    if scc[num1[i]]=scc[num2[i]] then exit(false);
  exit(true);
end;

begin
  into;
  if work then writeln('panda is telling the truth...')
    else writeln('the evil panda is lying again');
end.