POJ3207+tarjan+2-sat

 /*
 2-sat
 题意：给定一个圆，圆上一些点。两点一线。现给出一些线，这些线可以在圆内连起来，也可以在圆外。
     问有没有可能所有的线画完 且 不出现相交。
 思路：把线画在圆内或圆外 看成一个组合。其它的则建边。
 */
 #include<stdio.h>
 #include<string.h>
 #include<stdlib.h>
 #include<math.h>
 #include<algorithm>
 #include<iostream>
 #include<set>
 #include<stack>
 #include<queue>
 #include<map>
 #include<vector>
 using namespace std;

 typedef long long int64;
 //typedef __int64 int64;
 typedef pair<int64,int64> PII;
 #define MP(a,b) make_pair((a),(b))
 const int maxn = ;
 const int maxm = ;
 const int inf = 0x7fffffff;
 const double pi=acos(-1.0);
 const double eps = 1e-;

 struct Edge{
     int u,v,next;
 }edge[ maxm*+ ];
 int cnt,head[ maxn ];
 int vis[ maxn ];
 int dfn[ maxn ],low[ maxn ];
 int belong[ maxn ],Cnt,id;
 stack<int>s;
 struct EDGE{
     int l,r;
 }E[ maxn ];

 void init(){
     id = Cnt = ;
     cnt = ;
     while( !s.empty() )
         s.pop();
     memset( head,-,sizeof( head ) );
     memset( vis,-,sizeof( vis ) );
     memset( dfn,-,sizeof( dfn ) );
     memset( low,-,sizeof( low ) );
 }

 void addedge( int a,int b ){
     edge[ cnt ].u = a;
     edge[ cnt ].v = b;
     edge[ cnt ].next = head[ a ];
     head[ a ] = cnt++;
 }

 bool ok( int L,int R ){
     int x1 = E[L].l;
     int y1 = E[L].r;
     int x2 = E[R].l;
     int y2 = E[R].r;
     if( x2>x1&&x2<y1 ){
         if( y2>=y1 ) return true;
         if( y2<=x1 ) return true;
     }
     if( y2>x1&&y2<y1 ){
         if( x2>=y1 ) return true;
         if( x2<=x1 ) return true;
     }
     return false;
 }

 void tarjan( int cur ){
     dfn[ cur ] = low[ cur ] = id++;
     vis[ cur ] = ;
     s.push( cur );
     for( int i=head[ cur ];i!=-;i=edge[i].next ){
         int nxt = edge[ i ].v;
         if( dfn[ nxt ]==- ){
             tarjan( nxt );
             low[ cur ] = min( low[ cur ],low[ nxt ] );
         }
         else if( vis[ nxt ]== ){
             low[ cur ] = min( low[ cur ],dfn[ nxt ] );
         }
     }
     if( dfn[ cur ]==low[ cur ] ){
         Cnt ++;
         while(  ){
             int tmp = s.top();
             s.pop();
             vis[ tmp ] = ;
             belong[ tmp ] = Cnt;
             if( tmp==cur ) break;
         }
     }
 }

 int main(){
     int n,m;
     while( scanf("%d%d",&n,&m)== ){
         init();
         int a,b;
         for( int i=;i<=m;i++ ){
             scanf("%d%d",&a,&b);
             a++,b++;
             E[ i ].l = min( a,b );
             E[ i ].r = max( a,b );
         }
         for( int i=;i<=m;i++ ){
             for( int j=i+;j<=m;j++ ){
                 if( ok( i,j )==true ){
                     addedge( i,j+m );
                     addedge( j+m,i );
                     addedge( i+m,j );
                     addedge( j,i+m );
                 }
             }
         }//build mat
         for( int i=;i<=*m;i++ ){
             if( dfn[i]==- ){
                 tarjan( i );
             }
         }
         //
         bool f = true;
         for( int i=;i<=m;i++ ){
             if( belong[i]==belong[i+m] ){
                 f = false;
                 break;
             }
         }
         if( f==false ) printf("the evil panda is lying again");
         else printf("panda is telling the truth...");
         printf("\n");
     }
     return ;
 }