hdu3253 Diamonds

hdu3253 Diamonds

可以用以下方法构造出一组解，从第n个菱形开始依次放置，放置第i个菱形时必须覆盖p1,p2,...,pi这些点且被第i+1个菱形包含
此时菱形中点的可行区域其实是一个特殊的半平面交（只有4种方向的半平面），呵呵~
我用了4个堆去维护这四个半平面，时间复杂度为O(nlogn)
找可行区域的整点时要注意一下，不能出现0.5什么的

#include  < iostream >
#include  < cstdlib >
#include  < cstdio >

using   namespace  std;

int  n,x[ 210 ],y[ 210 ],d[ 210 ],rx[ 210 ],ry[ 210 ];

struct  heap
{
   int  t1,t2;
   bool  cmp( const   int   & u, const   int   & v)
  {  return  t1 * x[u] + t2 * y[u] > t1 * x[v] + t2 * y[v]; }
   int  n,h[ 210 ],e[ 210 ];
   void  del( int  i)
  {
    e[h[i]] = 0 ;
     if  (i == n)
    { n -- ;  return ; }
    h[i] = h[n -- ]; e[h[i]] = i;
    down(i); up(i);
  }
   void  build( void )
  {
     for ( int  i = (n >> 1 );i >= 1 ;i -- )
      down(i);
  }
   void  up( int  i)
  {
     int  t,j;
     while (i > 1 )
    {
      j = (i >> 1 );
       if  (cmp(h[i],h[j]))
      { 
        t = h[i]; h[i] = h[j]; h[j] = t;
        e[h[i]] = i; e[h[j]] = j;
      }
       else   break ;
      i = j;
    }
  }
   void  down( int  i)
  {
     int  t,j;
     while ((j = (i << 1 )) <= n)
    {
       if  (j < n && cmp(h[j + 1 ],h[j])) j ++ ;
       if  (cmp(h[j],h[i]))
      { 
        t = h[i]; h[i] = h[j]; h[j] = t;
        e[h[i]] = i; e[h[j]] = j;
      }
       else   break ;
      i = j;
    }
  }
}h1,h2,h3,h4;

int  main( void )
{
   int  u,c,i,l1,r1,l2,r2;
  scanf( " %d " , & c);
   for (u = 1 ;u <= c;u ++ )
  {
    scanf( " %d " , & n);
    h1.n = h2.n = h3.n = h4.n = n;
     for (i = 1 ;i <= n;i ++ )
    {
      scanf( " %d%d%d " ,x + i,y + i,d + i);
      h1.h[i] = h1.e[i] = i;
      h2.h[i] = h2.e[i] = i;
      h3.h[i] = h3.e[i] = i;
      h4.h[i] = h4.e[i] = i;
    }
    h1.t1 = 1 ; h1.t2 = 1 ;
    h2.t1 =- 1 ; h2.t2 = 1 ;
    h3.t1 =- 1 ; h3.t2 =- 1 ;
    h4.t1 = 1 ; h4.t2 =- 1 ;
    h1.build();
    h2.build();
    h3.build();
    h4.build();
     for (i = n;i >= 1 ;i -- )
    {
      l1 = x[h1.h[ 1 ]] + y[h1.h[ 1 ]] - d[i];
      r1 = x[h3.h[ 1 ]] + y[h3.h[ 1 ]] + d[i];
      l2 =- x[h2.h[ 1 ]] + y[h2.h[ 1 ]] - d[i];
      r2 =- x[h4.h[ 1 ]] + y[h4.h[ 1 ]] + d[i];
       if  (i < n)
      {
        r1 = min(r1,rx[i + 1 ] + ry[i + 1 ] + d[i + 1 ] - d[i]);
        l1 = max(l1,rx[i + 1 ] + ry[i + 1 ] - d[i + 1 ] + d[i]);
        r2 = min(r2, - rx[i + 1 ] + ry[i + 1 ] + d[i + 1 ] - d[i]);
        l2 = max(l2, - rx[i + 1 ] + ry[i + 1 ] - d[i + 1 ] + d[i]);
      }
       if  ((l1 + l2) % 2 != 0 )
         if  (l1 < r1)
          l1 ++ ;
         else  l2 ++ ;
      rx[i] = (l1 - l2) / 2 ;
      ry[i] = (l1 + l2) / 2 ;
      h1.del(h1.e[i]);
      h2.del(h2.e[i]);
      h3.del(h3.e[i]);
      h4.del(h4.e[i]);
    }
    printf( " Case %d:\n " ,u);
     for (i = 1 ;i <= n;i ++ )
      printf( " %d %d\n " ,rx[i],ry[i]);
  }
   return   0 ;
}