POJ-3207 Ikki's Story IV - Panda's Trick 2sat

　　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 ;
 }