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
      
       0 do begin    too:=edge[i].toward;    if dfn[too]=0 then begin      tarjan(too);      if low[too]
       
        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.