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 mlines each contain two integers ai and bi, 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”.
Sample Input
4 2
0 1
3 2
Sample Output
panda is telling the truth...
Source
POJ Monthly--2007.03.04, Ikki
题目大意：
有一个圆圈，要在上面连线。一条线可以画在圆的外面，也可以画在圆的里面。给出m条线段，求这些线段能否两两不相交。
思路：
把一条线拆成两种可能，连在外面和连在里面，计作点i和点i'。输入的时候判断每两条线段的关系，然后苦逼地如下枚举：
如果两条边i和j必须一个画在内部，一个画在外部（判断是否有重叠的部分）
那么连边：
i->j’,
j->’，
I’->j，
J’->i，
因为只要求我们判断是否有解，tarjan偷懒大法。
源代码/pas:
type
point=record
x,y,next:longint;
end;
var
n,m,t,maxE:longint;
e:array[0..2000001]of point;
ls,low,dfn,s,comp:array[0..1000]of longint;
v:array[0..1000]of boolean;
function min(a,b:longint):longint;
begin
min:=a;
if b<a then
min:=b;
end;
procedure add(x,y:Longint);
begin
inc(maxE);
e[maxE].x:=x;
e[maxE].y:=y;
e[maxE].next:=ls[x];
ls[x]:=maxE;
End;
procedure tarjan(x:longint);
var
i,y:longint;
begin
inc(t);
dfn[x]:=t;
low[x]:=t;
inc(s[0]);
s[s[0]]:=x;
v[x]:=true;
i:=ls[x];
while i>0 do
begin
y:=e[i].y;
if dfn[y]=0 then
begin
tarjan(y);
low[x]:=min(low[y],low[x]);
end
else
if v[y] then
low[x]:=min(low[x],dfn[y]);
i:=e[i].next;
end;
if dfn[x]=low[x] then
begin
inc(comp[0]);
repeat
y:=s[s[0]];
comp[y]:=comp[0];
v[y]:=false;
dec(s[0]);
until x=y;
end;
end;
procedure check;
var
i:longint;
begin
for i:=1 to n*2 do
if dfn[i]=0 then
tarjan(i);
for i:=1 to n do
if comp[i]=comp[i+n] then
begin
writeln('the evil panda is lying again');
halt;
End;
writeln('panda is telling the truth...');
End;
procedure main;
var
i,j:longint;
x,y:array[1..1000]of longint;
begin
fillchar(x,sizeof(x),0);
fillchar(y,sizeof(y),0);
readln(m,n);
for i:=1 to n do
begin
read(x[i],y[i]);
inc(x[i]);
inc(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:=1 to n-1 do
for j:=i+1 to n do
if ((x[i]<=x[j])and(y[i]>=x[j])and(y[i]<=y[j]))
or((x[i]>=x[j])and(x[i]<=y[j])and(y[i]>=y[j])) then
begin
add(i,j+n);
add(j,i+n);
add(i+n,j);
add(j+n,i);
end;
check;
end;
begin
main;
end.来源：https://www.cnblogs.com/olahiuj/p/5781308.html