bzoj 2850: 巧克力王国 K-D树

题目大意

http://www.lydsy.com/JudgeOnline/problem.php?id=2850

题解

对于每个人,我们发现它能够接受的巧克力中

如果对参数分别讨论,那么一定是一个连续的区间

所以我们利用K-D划分维度,然后直接搜就好了

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
inline void read(ll &x){
	x=0;char ch;bool flag = false;
	while(ch=getchar(),ch<'!');if(ch == '-') ch=getchar(),flag = true;
	while(x=10*x+ch-'0',ch=getchar(),ch>'!');if(flag) x=-x;
}
inline ll min(ll a,ll b,ll c){return min(a,min(b,c));}
inline ll max(ll a,ll b,ll c){return max(a,max(b,c));}
const int maxn = 50010;
const int dem = 2;
struct Node{
	ll pos[2],sum,val;
	ll minn[2],maxx[2];
	Node *ch[2];
	void update(){
		minn[0] = min(pos[0],ch[0]->minn[0],ch[1]->minn[0]);
		maxx[0] = max(pos[0],ch[0]->maxx[0],ch[1]->maxx[0]);
		minn[1] = min(pos[1],ch[0]->minn[1],ch[1]->minn[1]);
		maxx[1] = max(pos[1],ch[0]->maxx[1],ch[1]->maxx[1]);
		sum = val + ch[0]->sum + ch[1]->sum;
	}
}*null,*root;
Node T[maxn],aNoUseNode;
inline void init(){
	null = &aNoUseNode;null->ch[0] = null->ch[1] = null;
	null->sum = null->val = 0;
	null->minn[0] = 1LL<<60;null->maxx[0] = -(1LL<<60);
	null->minn[1] = 1LL<<60;null->maxx[1] = -(1LL<<60);
}
int split[maxn],now;
inline bool cmp(const Node &a,const Node &b){
	return a.pos[split[now]] < b.pos[split[now]];
}
Node *build(int l,int r,int s){
	if(l > r) return null;
	int mid = (l+r) >> 1;
	split[now = mid] = s % dem;
	nth_element(T+l,T+mid,T+r+1,cmp);
	Node *p = &T[mid];
	p->ch[0] = build(l,mid-1,s+1);
	p->ch[1] = build(mid+1,r,s+1);
	p->update();return p;
}
ll a,b,c;
inline bool judge1(Node *p){
	return (a*p->minn[0] + b*p->minn[1] < c)
	&& (a*p->minn[0] + b*p->maxx[1] < c)
	&& (a*p->maxx[0] + b*p->minn[1] < c)
	&& (a*p->maxx[0] + b*p->maxx[1] < c);
}
inline bool judge2(Node *p){
	return (a*p->minn[0] + b*p->minn[1] >= c)
	&& (a*p->minn[0] + b*p->maxx[1] >= c)
	&& (a*p->maxx[0] + b*p->minn[1] >= c)
	&& (a*p->maxx[0] + b*p->maxx[1] >= c);
}
ll query(Node *p){
	if(p == null) return 0;
	ll ret = 0;
	if(a*p->pos[0] + b*p->pos[1] < c) ret += p->val;
	if(judge1(p->ch[0])) ret += p->ch[0]->sum;
	else if(!judge2(p->ch[0])) ret += query(p->ch[0]);
	if(judge1(p->ch[1])) ret += p->ch[1]->sum;
	else if(!judge2(p->ch[1])) ret += query(p->ch[1]);
	return ret;
}
int main(){
	init();
	ll n,m;read(n);read(m);
	for(int i=1;i<=n;++i){
		read(T[i].pos[0]);read(T[i].pos[1]);
		read(T[i].val);T[i].ch[0] = T[i].ch[1] = null;
		T[i].update();
	}
	root = build(1,n,1);
	while(m--){
		read(a);read(b);read(c);
		printf("%lld\n",query(root));
	}
	getchar();getchar();
	return 0;
}