poj 1151 Atlantis

用离散化处理就可以了：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#include<cstring>
#include<vector>
#include<string>
#define LL long long
using namespace std;
class Point
{
public:
    double x1,y1,x2,y2;    
}p[124];
double tx[224];
bool cmp1( double a, double b )
{
    return a < b;    
} 
bool cmp2( Point a , Point b )
{
    return a.y1 < b.y1;    
}
double Solve( int N ,int cnt )
{
    double s = 0;
    sort( tx , tx + cnt , cmp1 );
    sort( p , p + N , cmp2 );
    cnt = unique( tx, tx + cnt  ) - tx;
    for( int i = 1 ; i <cnt ; i ++ )
    {
         double l = tx[i-1],r = tx[i],up = -1.0,down = -1.0;
         for( int j = 0 ; j < N; j ++ )
         {
              if( l >= p[j].x1 && r <= p[j].x2  )
              {
                  if( p[j].y1 > up )
                  {
                      s += ( r - l )*( up - down );
                      up = p[j].y2 ; down = p[j].y1;        
                  }        
                  else if( up < p[j].y2 )
                           up = p[j].y2;
              }              
//              printf( "s = %lf %lf %lf\n",s,l,r );
         }     
         s += ( r - l )*( up - down );
    }
    return s;
}
int main(  )
{
    int Case = 1,N;
    while( scanf( "%d",&N ),N )
    {
         int cnt = 0;
         for( int i = 0 ; i < N; i ++ )
         {
             scanf( "%lf%lf%lf%lf",&p[i].x1,&p[i].y1,&p[i].x2,&p[i].y2 );
            tx[cnt++] = p[i].x1;
            tx[cnt++] = p[i].x2;
         }
         printf( "Test case #%d\n",Case++ );
         printf( "Total explored area: %.2f\n\n",Solve( N ,cnt) );    
    }
    //system( "pause" );
    return 0;
}