poj3207 Ikki's Story IV - Panda's Trick

Description

liympanda, one of Ikki’s friend, likes playing games with Ikki. Today after minesweeping with Ikki and winning so many times, he is tired of such easy games and wants to play another game with Ikki.

liympanda has a magic circle and he puts it on a plane, there are n points on its boundary in circular border: 0, 1, 2, …, n − 1. Evil panda claims that he is connecting m pairs of points. To connect two points, liympanda either places the link entirely inside the circle or entirely outside the circle. Now liympanda tells Ikki no two links touch inside/outside the circle, except on the boundary. He wants Ikki to figure out whether this is possible…

Despaired at the minesweeping game just played, Ikki is totally at a loss, so he decides to write a program to help him.

Input

The input contains exactly one test case.

In the test case there will be a line consisting of of two integers: n and m (n ≤ 1,000, m ≤ 500). The following m lines each contain two integers a_i and b_i, which denote the endpoints of the ith wire. Every point will have at most one link.

Output

Output a line, either “panda is telling the truth...” or “the evil panda is lying again”.

题解：

这一题的主要就是建图（2-SAT），我想先把把每条边都抽象成一个点a，再把a点拆成两个点，分别为a0和a1，分别表示这条边从内部连何从外部连。

代码（一部分）：

function check(i,j:longint):boolean;

  begin

    if(x[j]>x[i])and(x[j]<y[i])and((y[j]<x[i])or(y[j]>y[i]))then exit(true);

    if (y[j]>x[i])and(y[j]<y[i])and((x[j]<x[i])or(x[j]>y[i]))then exit(true);

    check:=false;

  end;

procedureadd(x,y:longint);

  begin

    inc(e);

    side[e].x:=x;

    side[e].y:=y;

    side[e].next:=last[x];

    last[x]:=e;

  end;

procedure init;

var i,j:longint;

begin

  readln(n,m);

  for i:=1to m do

    begin

      readln(x[i],y[i]);

      if x[i]>y[i]then

        begin

          x[i]:=x[i] xor y[i];

          y[i]:=x[i] xor y[i];

          x[i]:=x[i] xor y[i];

        end;

    end;

  for i:=1to m-1 do

    for j:=i+1 to m do

      if check(i,j)then

        begin

          add(i,j+m);

          add(i+m,j);

          add(j,i+m);

          add(j+m,i);

        end;

end;