poj 3304 Segments

这个题就是纯暴力枚举：枚举所有顶点的连线的直线与所有线段是否有交点；

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#include<cstring>
#include<vector>
using namespace std;
class Line
{
public:
      double x1,x2,y1,y2;
}line[124],L[4];
int dcmp( double x )
{
   if( fabs( x ) < 1.0e-8 ) return 0;
   if( x < 0 ) return -1;
   return 1;    
}
double multi( double x1 ,double x2 ,double y1 ,double y2 )
{
    return x1 * y2 - x2 * y1;    
}
int segment( Line a , Line b )
{
   int d1 = dcmp( multi( a.x1 - b.x1 , a.x2 - b.x1 ,a.y1 - b.y1,a.y2 - b.y1 ) );
   int d2 = dcmp( multi( a.x1 - b.x2 , a.x2 - b.x2 ,a.y1 - b.y2,a.y2 - b.y2 ) );
   return d1 * d2;    
}
bool Solve( int n )
{
    for( int i = 0 ; i < n ; i ++ )
    {
       for( int j = 0 ; j < n  ; j ++  )
       {
          int cnt = 0;
          if( dcmp(line[i].x1 - line[j].x1) || dcmp( line[i].y1 - line[j].y1 ))
          {
              L[cnt].x1 = line[i].x1;L[cnt].y1 = line[i].y1;
              L[cnt].x2 = line[j].x1;L[cnt++].y2 = line[j].y1;
          }
          if( dcmp(line[i].x1 - line[j].x2) || dcmp( line[i].y1 - line[j].y2 ) )
          {
              L[cnt].x1 = line[i].x1;L[cnt].y1 = line[i].y1;
              L[cnt].x2 = line[j].x2;L[cnt++].y2 = line[j].y2;
          }
          if( dcmp(line[i].x2 - line[j].x1) || dcmp( line[i].y2 - line[j].y1 ) )
          {
              L[cnt].x1 = line[i].x2;L[cnt].y1 = line[i].y2;
              L[cnt].x2 = line[j].x1;L[cnt++].y2 = line[j].y1;
          }
          if(dcmp(line[i].x2 - line[j].x2) || dcmp( line[i].y2 - line[j].y2 ))
          {
              L[cnt].x1 = line[i].x2;L[cnt].y1 = line[i].y2;
              L[cnt].x2 = line[j].x2;L[cnt++].y2 = line[j].y2;
          }
          for( int l = 0 ; l < cnt ; l ++ )
          {
              int k;
              for( k = 0; k < n ; k ++ )
              {
                  if( segment(L[l] , line[k]) > 0 )
                    break;        
              }    
              if( k == n ) return true;
          }    
       }
    }    
    return false;
}
int main(  )
{
    int N,n;
    while( scanf( "%d",&N )==1 )
    {
      while( N-- )
      {
           scanf( "%d",&n );
           for( int i = 0 ; i < n ; i++ )
           {
               scanf( "%lf %lf %lf %lf",&line[i].x1,&line[i].y1,&line[i].x2,&line[i].y2 );        
           }    
           if( Solve( n ) ) printf( "Yes!\n" );
           else printf( "No!\n" );
      }
    }
    //system( "pause" );
    return 0;
}