BZOJ2850 Chocolate Kingdom

Test instructions: Given a heap of points, each point has the right value, each time in the line $ax + by + C = 0$ The weight of the point and

KD Tree maintain the point right in the two-dimensional interval and just fine ... Build complexity $o (n * logn) $, one-time query time $o (\sqrt{n}) $

1 /**************************************************************2 problem:28503 User:rausen4 language:c++5 result:accepted6 time:15568 Ms7 memory:4220 KB8 ****************************************************************/9  Ten#include <cstdio> One#include <algorithm> A   - using namespacestd; -typedefLong Longll; the Const intN = 5e4 +5; -   -Inlineintread (); -   + intN, M; - intA, B, C, D; +   A structPoint { at     intx[2], V; -       - Point () {} -       -Inlinevoid Get() { -x[0] = Read (), x[1] = read (), V =read (); in     } -       to     int&operator[] (inti) { +         returnX[i]; -     } theInlineBOOL operator< (ConstPoint &p)Const { *         returnX[D] <P.x[d]; $     }Panax Notoginseng } P[n]; -   theInlineBOOLCheckintXinty) { +     returnA * x + B * Y <C; A } the   + structKd_tree { -Kd_tree *ls, *rs; $ Point p; $     intmn[2], mx[2]; - ll sum; -       the Kd_tree () { -ls = rs =NULL;Wuyip =Point (); thesum =0; -     } Wu       -     #defineLen (1 << 16) AboutInlinevoid*operator New(size_t) { $         StaticKd_tree *mempool, *C; -         if(Mempool = =c) -Mempool = (c =NewKd_tree[len]) +Len; -*c =Kd_tree (); A         returnC++; +     } the     #undefLen -       $InlineintCalc () { the         returnCheck (mn[0], mn[1]) + Check (mn[0], mx[1]) + Check (mx[0], mn[1]) + Check (mx[0], mx[1]); the     } the       theInlinevoidUpdate () { -         Static inti; in          for(i =0; I <2; ++i) { theMn[i] = Mx[i] =P[i]; the             if(LS) { AboutMn[i] = min (mn[i], LS-mn[i]); theMx[i] = max (mx[i], LS-mx[i]); the             } the             if(RS) { +Mn[i] = min (Mn[i], RS-mn[i]); -Mx[i] = max (Mx[i], RS-mx[i]); the             }Bayi         } thesum =p.v; the         if(ls) sum + = lssum; -         if(rs) Sum + = rs-sum; -     } the        the     #defineMid (L + R >> 1) the     voidBuildintLintRintNow, point *P) { theD =Now ; -Nth_element (P + L, p + Mid, p + r +1); thep =P[mid]; the         if(L <mid) { thels =New() Kd_tree;94LS-to build (L, mid-1, !Now , P); the         } the         if(Mid <r) { thers =New() Kd_tree;98RS-Build (Mid +1R!Now , P); About         } - update ();101     }102     #undefMid103      104 ll query () { thell res =0;106         intcntl, cntr;107         if(Check (p[0], p[1]) Res + =p.v;108         if(LS) {109CNTL = ls-Calc (); the             if(Cntl = =4) Res + = ls-sum;111             Else if(cntl) Res + = ls-query (); the         }113         if(RS) { theCntr = RSCalc (); the             if(Cntr = =4) Res + = RS-sum; the             Else if(cntr) Res + = RS-query ();117         }118         returnRes;119     } -} *T;121  122 intMain () {123     inti;124n = Read (), M =read (); the      for(i =1; I <= N; ++i)126P[i].Get();127T =New() Kd_tree; -T-Build (1N0, p);129      while(m--) { theA = Read (), B = Read (), C =read ();131printf"%lld\n", T-query ()); the     }133     return 0;134 }135  136InlineintRead () {137     Static intx, SGN;138     Static Charch;139x =0, SGN =1, ch =GetChar (); $      while(Ch <'0'||'9'<ch) {141         if(ch = ='-') SGN =-1;142CH =GetChar ();143     }144      while('0'<= CH && Ch <='9') {145x = x *Ten+ CH-'0';146CH =GetChar ();147     }148     returnSGN *x;149}

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BZOJ2850 Chocolate Kingdom