[二进制分组 线段树 || 点分治 分治] UOJ #191 【集训队互测2016】Unknown

详见lzz的集训队论文

二进制分组做法

二进制分组是在线段树的结构上做的 方便区间查询
至于删除 采用延迟重构的思想 每一层只有最后一个区间是萎的 我们需要递归下去 询问还是O(logn)个节点 重构复杂度势能分析下O(nlogn)
只有上凸包是有效的 合并的时候采用归并加Graham可以做到O(n) 不然以我的常数 T的血惨
但是卡内存 只有90分

#include<cstdio>#include<cstdlib>#include<algorithm>#include<vector>#include<cmath>#include<cassert>#define pb push_backusing namespace std;typedef double ld;typedef long long ll;inline char nc(){  static char buf[100000],*p1=buf,*p2=buf;  return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline void read(int &x){  char c=nc(),b=1;  for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;  for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}const int N=550005;const ld PI=acos(-1.0);struct PP{  ll x,y;  PP(ll x=0,ll y=0):x(x),y(y) { }  friend PP operator + (PP A,PP B){ return PP(A.x+B.x,A.y+B.y); }  friend PP operator - (PP A,PP B){ return PP(A.x-B.x,A.y-B.y); }  friend ll operator * (PP A,PP B){ return A.x*B.y-A.y*B.x; }  friend ld Ang(PP A,PP B){    ld ang=atan2(B.y-A.y,B.x-A.x);    return (B.y<A.y&&B.x<=A.x)?ang+2*PI:ang;  }  bool operator < (const PP &B) const{    return x==B.x?y<B.y:x<B.x;  }};vector<int> hull[N<<1];int tmp[N]; int cnt,pnt;PP pp[N];int ncnt;int last[N];int ls[N<<1],rs[N<<1],lp[N<<1],lb[N<<1],rb[N<<1];int tag[N<<1];inline void BH(int x,int l,int r){  hull[x].clear();  if (l==r) { hull[x].pb(l); return; }  int p=0,q=0; cnt=0;  //assert(tag[ls[x]] && tag[rs[x]]);  for (;p<hull[ls[x]].size() || q<hull[rs[x]].size();){    int t;    if (p==hull[ls[x]].size())      t=hull[rs[x]][q++];    else if (q==hull[rs[x]].size())      t=hull[ls[x]][p++];    else if (pp[hull[ls[x]][p]].x<pp[hull[rs[x]][q]].x)      t=hull[ls[x]][p++];    else      t=hull[rs[x]][q++];    if (cnt && pp[t].x==pp[tmp[cnt]].x){      if (pp[tmp[cnt]].y<pp[t].y) tmp[cnt]=t;    }else      tmp[++cnt]=t;  }  pnt=0;  for (int i=1;i<=cnt;i++){    while (pnt>=2 && (pp[tmp[i]]-pp[tmp[pnt]])*(pp[tmp[pnt]]-pp[tmp[pnt-1]])<=0) pnt--;    tmp[++pnt]=tmp[i];  }  for (int i=1;i<=pnt;i++)    hull[x].pb(tmp[i]);}inline ll query(int x,PP p){  int L=-1,R=hull[x].size()-1,MID;  while (L+1<R){    MID=(L+R)>>1;    if ((pp[hull[x][MID+1]]-pp[hull[x][MID]])*p<=0)      L=MID;    else      R=MID;  }  return p*pp[hull[x][R]];}inline void Build(int &x,int l,int r,int d=0){  x=++ncnt; lp[x]=last[d]; last[d]=x; lb[x]=l; rb[x]=r;  if (l==r) return;  int mid=(l+r)>>1;  Build(ls[x],l,mid,d+1); Build(rs[x],mid+1,r,d+1);}inline void Add(int x,int l,int r,int t){  if (t==r && lp[x]) tag[lp[x]]=1,BH(lp[x],lb[lp[x]],rb[lp[x]]);  if (l==r) return;  int mid=(l+r)>>1;  if (t<=mid) Add(ls[x],l,mid,t);  else Add(rs[x],mid+1,r,t);}inline void Del(int x,int l,int r,int t){  tag[x]=0;  if (l==r) return;  int mid=(l+r)>>1;  if (t<=mid) Del(ls[x],l,mid,t);  else Del(rs[x],mid+1,r,t);}ll Ret;inline void Query(int x,int l,int r,int ql,int qr,PP p){  if (ql<=l && r<=qr && (tag[x] || l==r)){    if (l<r)      Ret=max(Ret,query(x,p));    else      Ret=max(Ret,p*pp[l]);    return;  }  int mid=(l+r)>>1;  if (ql<=mid) Query(ls[x],l,mid,ql,qr,p);  if (qr>mid) Query(rs[x],mid+1,r,ql,qr,p);}const int P=998244353;int main(){  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  int order,x,y,l,r;  int m; read(m);  while (1){    read(m); if (m==0) break;    int tot=0,rt; int n=1; while (n<m) n<<=1;    ncnt=0;    Build(rt,1,n);    int ans=0;    while (m--){      read(order);       if (order==1)    read(x),read(y),pp[++tot]=PP(x,y),Add(1,1,n,tot);      else if (order==2)    Del(1,1,n,tot--);      else{    Ret=-1LL<<60; read(l); read(r); read(x); read(y);    Query(1,1,n,l,r,PP(x,y));    ans^=((Ret%P)+P)%P;      }    }    printf("%d\n",ans);    for (int i=1;i<=ncnt;i++) ls[i]=rs[i]=lp[i]=last[i]=tag[i]=0,hull[i].clear();  }  return 0;}

操作树上点分治

发现加入删除就类似dfs中栈的过程 这样我们把操作树建出来就转化为一条方向到根的路径 点分后统计下重心到根的贡献更新下 这个再用一个分治就好了 具体最优能做到几个log 看论文 不然可能还是T的血惨
还是卡内存 好不容易卡进了 被cha了 只有97

#include<cstdio>#include<cstdlib>#include<algorithm>#include<vector>#define pb push_backusing namespace std;//typedef pair<int,int> abcd;#define abcd PP#define first x#define second ytypedef double ld;typedef long long ll;inline char nc(){  static char buf[100000],*p1=buf,*p2=buf;  return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline void read(int &x){  char c=nc(),b=1;  for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;  for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}const int N=300005;const ld PI=acos(-1.0);struct PP{  ll x,y;  PP(ll x=0,ll y=0):x(x),y(y) { }  friend PP operator + (PP A,PP B){ return PP(A.x+B.x,A.y+B.y); }  friend PP operator - (PP A,PP B){ return PP(A.x-B.x,A.y-B.y); }  friend ll operator * (PP A,PP B){ return A.x*B.y-A.y*B.x; }  bool operator < (const PP &B) const{    return x==B.x?y<B.y:x<B.x;  }};struct edge{  int v,next;}G[N<<1];int head[N],qhead[N],inum;inline void add(int u,int v,int p,int *head=::head){  /*G[p].u=u;*/ G[p].v=v; G[p].next=head[u]; head[u]=p;}#define V G[p].vint ncnt; //int Stack[N],Pnt;PP pp[N];int fat[N],depth[N];bool del[N];int minv=1<<30,size[N],sum,rt;inline void Root(int u,int fa){  size[u]=1; int maxv=0;  for (int p=head[u];p;p=G[p].next)    if (V!=fa && !del[V])      Root(V,u),size[u]+=size[V],maxv=max(maxv,size[V]);  maxv=max(maxv,sum-size[u]);  if (minv>maxv) minv=maxv,rt=u;}//vector<int> que[N];int tot,uu[N]/*,vv[N]*/; ll ans[N];PP qp[N];bool cmp(abcd x,abcd y){  return qp[x.second]*qp[y.second]<=0;}PP po[N]; int pcnt; PP tmpp[N];abcd qq[N]; int qcnt; //abcd tmp[N];inline void divide(int l,int r,int ql,int qr,int &hcnt){  if (l==r){    sort(qq+ql,qq+qr+1,cmp);    for (int i=ql;i<=qr;i++)      ans[qq[i].second]=max(ans[qq[i].second],qp[qq[i].second]*po[l]);    hcnt=1;    return;  }  int lh=0,rh=0; int mid=(l+r)>>1;  int lp=ql-1,rp=qr+1;  for (int i=ql;i<=qr;i++)    if (qq[i].first<=mid) tmpp[++lp]=qq[i]; else tmpp[--rp]=qq[i];  for (int i=ql;i<=qr;i++) qq[i]=tmpp[i];  divide(l,mid,ql,lp,lh);  divide(mid+1,r,rp,qr,rh);  int j=1;  for (int i=rp;i<=qr;i++){    while (j+1<=lh && qp[qq[i].second]*po[l+j-1]<qp[qq[i].second]*po[l+(j+1)-1])      j++;    ans[qq[i].second]=max(ans[qq[i].second],qp[qq[i].second]*po[l+j-1]);  }  if (!(l==1 && r==pcnt)){    int p=1,q=1; int cnt=0;    while (p<=lh || q<=rh){      PP t;      if (p==lh+1 || (q<=rh && po[mid+q].x<=po[l+p-1].x))    t=po[mid+(q++)];      else    t=po[l+(p++)-1];      if (!cnt || tmpp[cnt].x!=t.x)    tmpp[++cnt]=t;      else    tmpp[cnt].y=max(tmpp[cnt].y,t.y);    }    int pnt=0;    for (int i=1;i<=cnt;i++){      while (pnt>=2 && (tmpp[i]-tmpp[pnt])*(tmpp[pnt]-tmpp[pnt-1])<=0) pnt--;      tmpp[++pnt]=tmpp[i];    }    hcnt=pnt;    for (int i=1;i<=pnt;i++) po[l+i-1]=tmpp[i];    pnt=0,p=ql,q=rp;    while (p<=lp || q<=qr){      if (p==lp+1 || (q<=qr && qp[qq[q].second]*qp[qq[p].second]<0))    tmpp[++pnt]=qq[q++];      else    tmpp[++pnt]=qq[p++];    }    for (int i=1;i<=pnt;i++) qq[ql+i-1]=tmpp[i];  }}int R,Gg;inline void dfs(int u,int fa){  //for (int i:que[u])  for (int p=qhead[u];p;p=G[p].next){    int i=V;    if (depth[uu[i]]<=depth[Gg])      qq[++qcnt]=abcd(depth[Gg]-max(depth[uu[i]],depth[R])+1,i);  }  for (int p=head[u];p;p=G[p].next)    if (V!=fa && !del[V])      dfs(V,u);}inline void Divide(int u,int S){  sum=S; minv=1<<30; Root(u,0);  R=u; Gg=rt;  pcnt=0; int t=Gg; while (t!=R) po[++pcnt]=pp[t],t=fat[t]; po[++pcnt]=pp[R];  qcnt=0; dfs(Gg,fat[Gg]);  int tmp; if (qcnt) divide(1,pcnt,1,qcnt,tmp);  int tt=Gg;  del[Gg]=1;  if (Gg^R) Divide(u,S-size[Gg]);  for (int p=head[tt];p;p=G[p].next)    if (!del[V])      Divide(V,size[V]);}const int P=998244353;int main(){  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  int m; int order,l,r,x,y;  read(m);  while (1){    read(m); if (!m) break; int cur=0,Pnt=0,*Stack=size;tot=ncnt=0;    Stack[++Pnt]=++ncnt; cur=ncnt;    while (m--){      read(order);      if (order==1){    ++ncnt; if (cur) add(cur,ncnt,++inum),fat[ncnt]=cur;    depth[ncnt]=depth[fat[ncnt]]+1;    cur=ncnt; Stack[++Pnt]=ncnt;    read(x); read(y); pp[ncnt]=PP(x,y);      }else if (order==2){    cur=fat[cur]; Stack[Pnt--]=0;      }else if (order==3){    read(l); read(r); read(x); read(y);    ++tot; ans[tot]=-1LL<<60; qp[tot]=PP(x,y);  uu[tot]=Stack[l+1];    //vv[tot]=Stack[r+1]; que[vv[tot]].pb(tot);    //que[Stack[r+1]].pb(tot);    add(Stack[r+1],tot,++inum,qhead);      }    }    Divide(1,ncnt);    int Ans=0;    for (int i=1;i<=tot;i++)      Ans^=(ans[i]%P+P)%P;    printf("%d\n",Ans);    for (int i=1;i<=ncnt;i++) head[i]=del[i]=fat[i]=depth[i]=0,qhead[i]=0/*,que[i].clear()*/; inum=0;  }  return 0;}