bzoj2850 Chocolate Kingdom *

bzoj2850 Chocolate Kingdom

Test instructions

n chocolate, each with milk content, cocoa content and tasty value. M individuals, each with three weights a,b,c, if the milk content of a chocolate *a+ cocoa content *b<c can be accepted. Ask everyone to accept the sum of the delicious value of chocolate. n,m≤50000.

Exercises

For all chocolates, the tree node maintains the sum of the subtree's tasty values in addition to maintaining the maximum minimum value of the sub-tree's horizontal ordinate. At the time of the query if the value of the tree's milk content maximum by a + cocoa content maximum *b less than C is the whole subtrees tree can be accepted, otherwise as long as the subtree may have the opportunity to exist acceptable chocolate on the traversal of the subtrees tree.

Code:

1#include <cstdio>2#include <cstring>3#include <algorithm>4 #defineInc (I,J,K) for (int i=j;i<=k;i++)5 #defineMAXN 500106 #definell Long Long7 using namespacestd;8 9 inline ll read () {Ten     CharCh=getchar (); ll f=1, x=0; One      while(ch<'0'|| Ch>'9'){if(ch=='-') f=-1; Ch=GetChar ();} A      while(ch>='0'&&ch<='9') x=x*Ten+ch-'0', ch=GetChar (); -     returnf*x; - } the intN,m,f,rt; ll ans; - structp{intpos[2],v;BOOL operator< (ConstP &a)Const{returnpos[f]<a.pos[f];}} PS[MAXN]; - structNd{p POS;intmx[2],mn[2],LC,RC; ll SM;} NDS[MAXN]; - BOOLCheckintAintB,ll c,ll d,ll e) {returnc*a+d*b<e;} + voidUpdateintx) { -Inc (I,0,1){ +         if(NDS[X].LC) Ands[x].mx[i]=Max (Nds[x].mx[i],nds[nds[x].lc].mx[i]), atnds[x].mn[i]=min (nds[x].mn[i],nds[nds[x].lc].mn[i]); -         if(nds[x].rc) -nds[x].mx[i]=Max (Nds[x].mx[i],nds[nds[x].rc].mx[i]), -nds[x].mn[i]=min (nds[x].mn[i],nds[nds[x].rc].mn[i]); -     } -     if(NDS[X].LC) nds[x].sm+=Nds[nds[x].lc].sm; in     if(nds[x].rc) nds[x].sm+=Nds[nds[x].rc].sm; - } to intBuildintLintRintNow ) { +F=now;intMid= (l+r) >>1; Nth_element (ps+l,ps+mid,ps+r+1); -Inc (I,0,1) Nds[mid].mx[i]=nds[mid].mn[i]=ps[mid].pos[i]; NDS[MID].SM=PS[MID].V; nds[mid].pos=Ps[mid]; the     if(L<mid) Nds[mid].lc=build (l,mid-1, now^1);if(mid<r) Nds[mid].rc=build (mid+1, r,now^1); *Update (mid);returnmid; $ }Panax Notoginseng int Get(intx,ll a,ll b,ll c) { -     intq=0; theQ+=check (nds[x].mx[0],nds[x].mx[1],A,B,C); Q+=check (nds[x].mx[0],nds[x].mn[1],a,b,c); +Q+=check (nds[x].mn[0],nds[x].mx[1],A,B,C); Q+=check (nds[x].mn[0],nds[x].mn[1],a,b,c); A     returnQ; the } + voidQueryintx,ll a,ll b,ll c) { -     if(Check (nds[x].pos.pos[0],nds[x].pos.pos[1],A,B,C)) ans+=nds[x].pos.v;intDl=0, dr=0; $     if(NDS[X].LC) dl=Get(NDS[X].LC,A,B,C);if(nds[x].rc) dr=Get(nds[x].rc,a,b,c); $     if(dl==4) Ans+=nds[nds[x].lc].sm;Else if(DL) query (NDS[X].LC,A,B,C); -     if(dr==4) Ans+=nds[nds[x].rc].sm;Else if(DR) query (NDS[X].RC,A,B,C); - } the intMain () { -N=read (); M=read (); Inc (I,1, N) ps[i].pos[0]=read (), ps[i].pos[1]=read (), Ps[i].v=read (); Rt=build (1N0);WuyiInc (I,1, M) { thell X=read (), Y=read (), Z=read (); ans=0; Query (RT,X,Y,Z); printf"%lld\n", ans); -     } Wu     return 0; -}

20160906

bzoj2850 Chocolate Kingdom *