poj 1151 Atlantis

用离散化处理就可以了：

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#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;

}