题目链接：http://poj.org/problem?id=3207
题意：在一个圆圈上有n个点，现在用线把点两两连接起来，线只能在圈外或者圈内，现给出m个限制，第 i 个点和第 j 个点必须链接在一起，问是否存在可行解。
容易想到圈内和圈外分别表示2sat的两种状态，对每一个限制 i 和 j ，考虑所有其它横跨他们的限制，然后连边就可以了。
//STATUS:C++_AC_47MS_6300KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef long long LL;
typedef unsigned long long ULL;
//const
const int N=;
const int INF=0x3f3f3f3f;
const int MOD=,STA=;
const LL LNF=1LL<<;
const double EPS=1e-;
const double OO=1e15;
const int dx[]={-,,,};
const int dy[]={,,,-};
const int day[]={,,,,,,,,,,,,};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End int nod[N/][];
int first[N*],next[N*N],vis[N*N],S[N*];
int n,m,mt,cnt; struct Edge{
int u,v;
}e[N*N]; void adde(int a,int b)
{
e[mt].u=a,e[mt].v=b;
next[mt]=first[a];first[a]=mt++;
} int dfs(int u)
{
if(vis[u^])return ;
if(vis[u])return ;
int i;
vis[u]=;
S[cnt++]=u;
for(i=first[u];i!=-;i=next[i]){
if(!dfs(e[i].v))return ;
}
return ;
} int Twosat()
{
int i,j;
for(i=;i<n;i+=){
if(vis[i] || vis[i^])continue;
cnt=;
if(!dfs(i)){
while(cnt)vis[S[--cnt]]=;
if(!dfs(i^))return ;
}
}
return ;
} int main()
{
// freopen("in.txt","r",stdin);
int i,j,x,y;
while(~scanf("%d%d",&n,&m))
{
n<<=;
mem(first,-);mt=;
mem(vis,);
for(i=;i<m;i++){
scanf("%d%d",&nod[i][],&nod[i][]);
if(nod[i][]>nod[i][])swap(nod[i][],nod[i][]);
} for(i=;i<m;i++){
for(j=i+;j<m;j++){
if( (nod[j][]<nod[i][] && nod[j][]>nod[i][] && nod[j][]<nod[i][])
|| (nod[j][]>nod[i][] && nod[j][]<nod[i][] && nod[j][]>nod[i][])){
x=i<<,y=j<<;
adde(x,y^);
adde(x^,y);
adde(y,x^);
adde(y^,x);
}
}
} printf("%s\n",Twosat()?"panda is telling the truth...":"the evil panda is lying again"); }
return ;
}