bzoj 2850

比较基础的KD树。每个节点维护一个BOX，包含包含当当前子树的点的最小矩形，以及点权和，然后用“整个矩形都在直线的一侧”和“整个矩形都不在直线的一侧”剪枝。

 /**************************************************************
     Problem: 2850
     User: idy002
     Language: C++
     Result: Accepted
     Time:27240 ms
     Memory:3740 kb
 ****************************************************************/

 #include <cstdio>
 #include <algorithm>
 #define N 50010
 #define oo 1000000001
 #define fprintf(...)
 using namespace std;

 typedef long long dnt;

 struct Point {
     int x, y, h;
     void read() { scanf( "%d%d%d", &x, &y, &h ); }
     Point(){}
     Point( int x, int y, int h ):x(x),y(y),h(h){}
     bool in( dnt a, dnt b, int c ) {
         return a*x+b*y<c;
     }
 };
 typedef bool (*Cmp)( const Point &a, const Point &b );
 struct Box {
     int xmin, xmax;
     int ymin, ymax;
     dnt sum;
     Box():xmin(oo),xmax(-oo),ymin(oo),ymax(-oo){}
     void add( Point &p ) {
         xmin = min( xmin, p.x );
         xmax = max( xmax, p.x );
         ymin = min( ymin, p.y );
         ymax = max( ymax, p.y );
         sum += p.h;
     }
     void add( Box &b ) {
         xmin = min( xmin, b.xmin );
         xmax = max( xmax, b.xmax );
         ymin = min( ymin, b.ymin );
         ymax = max( ymax, b.ymax );
         sum += b.sum;
     }
     bool in( dnt a, dnt b, int c ) {
         return (a*xmin+b*ymin<c)
             && (a*xmax+b*ymin<c)
             && (a*xmin+b*ymax<c)
             && (a*xmax+b*ymax<c);
     }
     bool out( dnt a, dnt b, int c ) {
         return (a*xmin+b*ymin>=c)
             && (a*xmax+b*ymin>=c)
             && (a*xmin+b*ymax>=c)
             && (a*xmax+b*ymax>=c);
     }
 };
 struct Node {
     Point p;
     Box box;
     Cmp cmp;
     Node *ls, *rs;
 }pool[N], *tail=pool, *root;

 int n, m;
 Point pts[N];
 Cmp cmp[];

 bool cmpx( const Point &a, const Point &b ) {
     return a.x<b.x;
 }
 bool cmpy( const Point &a, const Point &b ) {
     return a.y<b.y;
 }
 Node *build( int lf, int rg, int c ) {
     if( lf>rg ) return ;
     Node *nd = ++tail;
     int mid=(lf+rg)>>;
     nth_element( pts+lf, pts+mid, pts+rg+, cmp[c] );
     nd->cmp = cmp[c];
     nd->p = pts[mid];
     nd->ls = build( lf, mid-, !c );
     nd->rs = build( mid+, rg, !c );
     if( nd->ls ) nd->box.add( nd->ls->box );
     if( nd->rs ) nd->box.add( nd->rs->box );
     nd->box.add( pts[mid] );
     fprintf( stderr, "(%d,%d,%d) ", pts[mid].x, pts[mid].y, pts[mid].h );
     return nd;
 }
 dnt query( Node *nd, int a, int b, int c ) {
     if( nd->box.in(a,b,c) ) return nd->box.sum;
     if( nd->box.out(a,b,c) ) return ;
     dnt rt = ;
     if( nd->ls ) rt += query( nd->ls, a, b, c );
     if( nd->rs ) rt += query( nd->rs, a, b, c );
     if( nd->p.in(a,b,c) ) rt += nd->p.h;
     return rt;
 }
 int main() {
     scanf( "%d%d", &n, &m );
     for( int i=; i<=n; i++ )
         pts[i].read();
     cmp[] = cmpx;
     cmp[] = cmpy;
     root = build( , n,  );
     fprintf( stderr, "\n" );
     for( int i=,a,b,c; i<=m; i++ ) {
         scanf( "%d%d%d", &a, &b, &c );
         printf( "%lld\n", query(root,a,b,c) );
     }
 }