BZOJ2716: [Violet 3]天使玩偶
题解: 通过取绝对值考虑后 发现是一个较复杂的cdq 突然考虑距离问题 用Kdtree写也是可行的 但是.....会退化 所以我们考虑用替罪羊树来防止退化
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <stack>
#include <queue>
#include <cmath>
#include <set>
#include <map>
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define link(x) for(edge *j=h[x];j;j=j->next)
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,r,l) for(int i=r;i>=l;i--)
const int MAXN=5e5+10;
const double eps=1e-8;
#define ll long long
const int inf=1e9+7;
#define alpha 0.75
using namespace std;
struct edge{int t,v;edge*next;}e[MAXN<<1],*h[MAXN],*o=e;
void add(int x,int y,int vul){o->t=y;o->v=vul;o->next=h[x];h[x]=o++;}
ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return x*f;
}
int d,rt;
typedef struct node{
int p[2],maxx[2],minn[2],c[2],sz;
friend bool operator<(node aa,node bb){
if(aa.p[d]==bb.p[d])return aa.p[d^1]<bb.p[d^1];
return aa.p[d]<bb.p[d];
}
}node;
node a[MAXN<<1];
int St[MAXN],tot,n,m,tot1;
node st[MAXN<<1];
int newnode(int t){
int x=St[tot--];
a[x].sz=1;
inc(i,0,1)a[x].p[i]=a[x].maxx[i]=a[x].minn[i]=st[t].p[i],a[x].c[i]=0;
return x;
}
void up(int x,int y){
if(!x||!y)return ;
inc(i,0,1)a[x].maxx[i]=max(a[x].maxx[i],a[y].maxx[i]),a[x].minn[i]=min(a[x].minn[i],a[y].minn[i]);
}
int built(int l,int r,int now){
int mid=(l+r)>>1;
d=now;nth_element(a+l,a+mid,a+r+1);
inc(i,0,1)a[mid].maxx[i]=a[mid].minn[i]=a[mid].p[i];
a[mid].c[0]=a[mid].c[1]=0;a[mid].sz=1;
if(l<mid)a[mid].c[0]=built(l,mid-1,now^1),up(mid,a[mid].c[0]);
if(r>mid)a[mid].c[1]=built(mid+1,r,now^1),up(mid,a[mid].c[1]);
a[mid].sz=a[a[mid].c[0]].sz+a[a[mid].c[1]].sz+1;
return mid;
}
int Built(int l,int r,int now){
int mid=(l+r)>>1;
d=now;nth_element(st+l,st+mid,st+r+1);
int x=newnode(mid);
if(l<mid)a[x].c[0]=Built(l,mid-1,now^1),up(x,a[x].c[0]);
if(r>mid)a[x].c[1]=Built(mid+1,r,now^1),up(x,a[x].c[1]);
a[x].sz=a[a[x].c[0]].sz+a[a[x].c[1]].sz+1;
return x;
}
void dfs(int x){
if(!x)return ;
dfs(a[x].c[0]);
st[++tot1]=a[x];St[++tot]=x;
dfs(a[x].c[1]);
}
void check(int &x,int now){
if(alpha*a[x].sz<a[a[x].c[0]].sz||alpha*a[x].sz<a[a[x].c[1]].sz){
tot=tot1=0;dfs(x);
x=Built(1,tot1,now);
}
}
void insert(int &x,int now){
if(!x){x=n;a[x].sz=1;return ;}
if(a[n].p[now]<a[x].p[now]){
insert(a[x].c[0],now^1);
}
else{
insert(a[x].c[1],now^1);
}
if(a[x].c[0])up(x,a[x].c[0]);
if(a[x].c[1])up(x,a[x].c[1]);
a[x].sz=a[a[x].c[0]].sz+a[a[x].c[1]].sz+1;
check(x,now);
}
node que;int ans;
int get_ans(node x,node y){
int ans1=0;
inc(i,0,1)ans1+=abs(x.p[i]-y.p[i]);
return ans1;
}
int getis(node x,node y){
int ans1=0;
inc(i,0,1){
if(y.p[i]<x.minn[i])ans1+=x.minn[i]-y.p[i];
if(y.p[i]>x.maxx[i])ans1+=y.p[i]-x.maxx[i];
}
return ans1;
}
void querty(int x){
if(!x)return ;
ans=min(get_ans(que,a[x]),ans);
int t1=(a[x].c[0]!=0)?getis(a[a[x].c[0]],que):inf;
int t2=(a[x].c[1]!=0)?getis(a[a[x].c[1]],que):inf;
if(t1<t2){
if(t1<ans)querty(a[x].c[0]);
if(t2<ans)querty(a[x].c[1]);
}
else{
if(t2<ans)querty(a[x].c[1]);
if(t1<ans)querty(a[x].c[0]);
}
}
int main(){
n=read();m=read();a[0].sz=0;
inc(i,1,n)a[i].p[0]=read(),a[i].p[1]=read();
rt=built(1,n,0);
int op;
while(m--){
op=read();
if(op==1){
n++;a[n].p[0]=read();a[n].p[1]=read();a[n].c[0]=a[n].c[1]=0;
inc(j,0,1)a[n].minn[j]=a[n].maxx[j]=a[n].p[j];
insert(rt,0);
}
else{
que.p[0]=read();que.p[1]=read();ans=inf;
querty(rt);
printf("%d\n",ans);
}
}
return 0;
}
2716: [Violet 3]天使玩偶
Time Limit: 80 Sec Memory Limit: 128 MBSubmit: 2744 Solved: 1184
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Description
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Input
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Output
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