luogu P4475 巧克力王国 KDT

LINK：P4475 巧克力王国

又复习了一遍KDT 这道题不需要 重构 所以没套替罪羊树.(神EI 说KDT在这道题上复杂度是假的 我不太懂

其实最重要的除了写法之外 还有一个比较重要的trick?

就是 查最远最近距离的时候的剪枝 其实我也有点忘了 大概乱写一下 只要正确 适时进行modify即可.

这道题 可以利用KDT 来优化暴力 因为 对于前面的限制可以看成二维平面上的点.

然后 每次查询查询一个矩形到当前点的最近最远距离 来进行剪枝即可.

注意开 long long

code

//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<ctime>
#include<cctype>
#include<queue>
#include<deque>
#include<stack>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cstring>
#include<string>
#include<ctime>
#include<cmath>
#include<cctype>
#include<cstdlib>
#include<queue>
#include<deque>
#include<stack>
#include<vector>
#include<algorithm>
#include<utility>
#include<bitset>
#include<set>
#include<map>
#define ll long long
#define db double
#define INF 1000000000
#define inf 1000000000000000ll
#define ldb long double
#define pb push_back
#define put_(x) printf("%d ",x);
#define get(x) x=read()
#define gt(x) scanf("%d",&x)
#define gi(x) scanf("%lf",&x)
#define put(x) printf("%d
",x)
#define putl(x) printf("%lld
",x)
#define rep(p,n,i) for(RE int i=p;i<=n;++i)
#define go(x) for(int i=lin[x],tn=ver[i];i;tn=ver[i=nex[i]])
#define fep(n,p,i) for(RE int i=n;i>=p;--i)
#define vep(p,n,i) for(RE int i=p;i<n;++i)
#define pii pair<int,int>
#define mk make_pair
#define RE register
#define P 13331ll
#define gf(x) scanf("%lf",&x)
#define pf(x) ((x)*(x))
#define uint unsigned long long
#define ui unsigned
#define EPS 1e-5
#define sq sqrt
#define S second
#define F first
#define l(x) t[x].l
#define r(x) t[x].r
#define sum(x) t[x].sum
#define v(x) t[x].v
#define mod 998244353
using namespace std;
char buf[1<<15],*fs,*ft;
inline char getc()
{
	return (fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<15,stdin),fs==ft))?0:*fs++;
}
inline int read()
{
	RE int x=0,f=1;RE char ch=getc();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getc();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getc();}
	return x*f;
}
inline ll Read()
{
	RE ll x=0,f=1;RE char ch=getc();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getc();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getc();}
	return x*f;
}
const int MAXN=50010,K=2;
int n,m,D,rt;
ll xx,yy,c,ans;
struct wy
{
	int l,r;ll sum,v;
	int d[K];
	int mx[K],mn[K];
}t[MAXN];
int q[MAXN];
inline int cmp(int a,int b){return t[a].d[D]<t[b].d[D];}
inline void pushup(int x)
{
	if(!l(x)&&!r(x))return;
	sum(x)+=sum(l(x))+sum(r(x));
	vep(0,K,i)
	{
		if(l(x))t[x].mx[i]=max(t[x].mx[i],t[l(x)].mx[i]),t[x].mn[i]=min(t[x].mn[i],t[l(x)].mn[i]);
		if(r(x))t[x].mx[i]=max(t[x].mx[i],t[r(x)].mx[i]),t[x].mn[i]=min(t[x].mn[i],t[r(x)].mn[i]);
	}
}
inline int build(int l,int r,int k)
{
	if(l>r)return 0;
	int mid=(l+r)>>1;D=k;
	nth_element(q+l,q+mid,q+r+1,cmp);
	int x=q[mid];sum(x)=v(x);
	vep(0,K,i)t[x].mx[i]=t[x].d[i],t[x].mn[i]=t[x].d[i];
	l(x)=build(l,mid-1,(k+1)%K);
	r(x)=build(mid+1,r,(k+1)%K);
	pushup(x);return x;
}
inline int check(int x)
{
	if(t[x].d[0]*xx+t[x].d[1]*yy>=c)return 0;
	return 1;
}
inline void ask(int x)
{
	ll maxx,minn;
	if(xx>0)maxx=t[x].mx[0]*xx,minn=t[x].mn[0]*xx;else maxx=t[x].mn[0]*xx,minn=t[x].mx[0]*xx;
	if(yy>0)maxx+=t[x].mx[1]*yy,minn+=t[x].mn[1]*yy;else maxx+=t[x].mn[1]*yy,minn+=t[x].mx[1]*yy;
	if(maxx<c){ans+=sum(x);return;}
	if(minn>=c)return;
	if(check(x))ans+=v(x);
	if(l(x))ask(l(x));
	if(r(x))ask(r(x));
}
int main()
{
	//freopen("1.in","r",stdin);
	get(n);get(m);
	rep(1,n,i)t[i].d[0]=read(),t[i].d[1]=read(),get(v(i)),q[i]=i;
	rt=build(1,n,0);
	rep(1,m,i)
	{
		get(xx),get(yy),get(c);
		ans=0;ask(rt);putl(ans);
	}
	return 0;
}