[模板]三维偏序

description

洛谷

data range

[nle 10^5,mle 2 imes 10^5 ]

kd-tree solution

直到现在才开始做三维偏序...还是用的(kd-tree)...

险险卡过.jpg

code

#include<bits/stdc++.h>
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<iomanip>
#include<cstring>
#include<complex>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<ctime>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define FILE "a"
#define Cpy(x,y) (memcpy(x,y,sizeof(x)))
#define Set(x,y) (memset(x,y,sizeof(x)))
#define mp make_pair
#define pb push_back
#define RG register
#define il inline
using namespace std;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef vector<int>VI;
typedef long long ll;
typedef double dd;
const dd eps=1e-10;
const int mod=1e9+7;
const int N=200010;
const int M=205;
const dd pi=acos(-1);
const int inf=2147483647;
const ll INF=1e18+1;
const ll P=100000;
il int read(){
	RG int data=0,w=1;RG char ch=getchar();
	while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
	if(ch=='-')w=-1,ch=getchar();
	while(ch<='9'&&ch>='0')data=(data<<3)+(data<<1)+(ch^48),ch=getchar();
	return data*w;
}

il void file(){
	srand(time(NULL)+rand());
	freopen(FILE".in","r",stdin);
	freopen(FILE".out","w",stdout);
}

int n,tot,rt,cd,f[N];
struct node{int d[3];}p[N];
bool operator <(node a,node b){return a.d[0]<b.d[0];}
struct kdnode{int d[2],l[2],r[2],s[2],v,sum;}t[N];
bool operator <(kdnode x,kdnode y){return x.d[cd]<y.d[cd];}
int l[2],r[2];

#define ls t[i].s[0]
#define rs t[i].s[1]
#define mid ((l+r)>>1)
il void update(int i){
	for(int k=0;k<2;k++){
		t[i].l[k]=t[i].r[k]=t[i].d[k];
		if(ls){
			t[i].l[k]=min(t[i].l[k],t[ls].l[k]);
			t[i].r[k]=max(t[i].r[k],t[ls].r[k]);
		}
		if(rs){
			t[i].l[k]=min(t[i].l[k],t[rs].l[k]);
			t[i].r[k]=max(t[i].r[k],t[rs].r[k]);			
		}
	}
	t[i].sum=t[ls].sum+t[rs].sum+t[i].v;
}

int rebuild(int l,int r,int D){
	if(l>r)return 0;
	cd=D;nth_element(t+l,t+mid,t+r+1);
	t[mid].s[0]=rebuild(l,mid-1,D^1);
	t[mid].s[1]=rebuild(mid+1,r,D^1);
	update(mid);return mid;
}

void insert(int &i,int D){
	if(!i){
		i=++tot;t[i].v=t[i].sum=0;
		t[i].l[0]=t[i].r[0]=t[i].d[0]=r[0];
		t[i].l[1]=t[i].r[1]=t[i].d[1]=r[1];
	}
	if(r[0]==t[i].d[0]&&r[1]==t[i].d[1])
	{t[i].v++;t[i].sum++;return;}
	insert(t[i].s[t[i].d[D]>r[D]],D^1);
	update(i);
}

#define ckm(dd) (l[dd]<=t[i].l[dd]&&t[i].r[dd]<=r[dd])
#define ckd(dd) (l[dd]<=t[i].d[dd]&&t[i].d[dd]<=r[dd])
#define cke(dd) (t[i].r[dd]<l[dd]||r[dd]<t[i].l[dd])
int query(int i,int D){
	if(!i||cke(0)||cke(1))return 0;
	if(ckm(0)&&ckm(1))return t[i].sum;
	RG int ret=0;
	if(ckd(0)&&ckd(1))ret+=t[i].v;
	ret+=query(ls,D^1)+query(rs,D^1);
	return ret;
}

int main()
{
	n=read();read();
	for(RG int i=1;i<=n;i++)a
		for(RG int j=0;j<3;j++)
			p[i].d[j]=read();
	sort(p+1,p+n+1);

	RG int pos=1,nxt=1,lim=10000,ans;
	while(pos<=n){
		while(nxt<=n&&(nxt==pos||p[nxt].d[0]==p[nxt-1].d[0])){
			r[0]=p[nxt].d[1];r[1]=p[nxt].d[2];insert(rt,0);nxt++;
		}
		for(RG int i=pos;i<nxt;i++){
			r[0]=p[i].d[1];r[1]=p[i].d[2];f[ans=query(rt,0)-1]++;
		}
		pos=nxt;
		if(pos<=n&&pos>=lim){
			rt=rebuild(1,tot,0);
			lim=pos/10000*10000+10000;
		}
	}
	for(RG int i=0;i<n;i++)printf("%d
",f[i]);
	return 0;
}

CDQ solution

现在才学(CDQ)..弱

code

不过(CDQ)的代码真的很短

#include<bits/stdc++.h>
#define RG register
#define il inline
using namespace std;
const int N=200010;
il int read(){
	RG int data=0,w=1;RG char ch=getchar();
	while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
	if(ch=='-')w=-1,ch=getchar();
	while(ch<='9'&&ch>='0')data=data*10+ch-48,ch=getchar();
	return data*w;
}
int n,m,k,ans[N],f[N],tr[N];struct node{int a,b,c,w,id;}p[N],tmp[N];
bool operator <(node x,node y){
	return x.a<y.a||(x.a==y.a&&(x.b<y.b||(x.b==y.b&&x.c<y.c)));
}
bool cmp1(node x,node y){return x.a==y.a&&x.b==y.b&&x.c==y.c;}
bool cmp2(node x,node y){return x.b<=y.b||(x.b==y.b&&x.c<=y.c);}
#define low(x) (x&(-x))
il void insert(int i,int v){for(;i<=k;i+=low(i))tr[i]+=v;}
il int query(int i){RG int r=0;for(;i;i-=low(i))r+=tr[i];return r;}
il void clear(int i){for(;i<=k;i+=low(i))tr[i]=0;}
void cdq(int l,int r){
	if(l==r)return;RG int mid=(l+r)>>1;cdq(l,mid);cdq(mid+1,r);
	for(RG int i=l,p1=l,p2=mid+1;i<=r;i++){
		if((p1<=mid&&cmp2(p[p1],p[p2]))||p2==r+1){
			insert(p[p1].c,p[p1].w);tmp[i]=p[p1++];
		}
		else{ans[p[p2].id]+=query(p[p2].c);tmp[i]=p[p2++];}
	}
	for(RG int i=l;i<=r;i++)clear(tmp[i].c),p[i]=tmp[i];
}
int main()
{
	n=read();k=read();
	for(RG int i=1;i<=n;i++){p[i].a=read();p[i].b=read();p[i].c=read();}
	sort(p+1,p+n+1);
	for(RG int i=1;i<=n;i++){
		p[++m]=p[i];p[m].w=1;p[m].id=m;
		while(i<n&&cmp1(p[i],p[i+1])){i++;p[m].w++;}
		ans[m]=p[m].w-1;
	}
	cdq(1,m);
	for(RG int i=1;i<=m;i++)f[ans[p[i].id]]+=p[i].w;
	for(RG int i=0;i<n;i++)printf("%d
",f[i]);
	return 0;
}