BZOJ2758 : [SCOI2012]Blinker的噩梦

首先将包含关系建树。

方法是将每个图形拆成上半边和下半边，从左往右扫描线，用Splay从下到上维护扫描线上所有图形。

每次加入一个新的图形$x$的时候，看看它下方第一个图形$y$，如果$y$是上半边，那么$x$的父亲就是$y$，否则是$y$的父亲。用同样的方法可以完成点定位。

然后每次相当于查询两点间的异或和，用树状数组维护dfs序即可。

时间复杂度$O((n+m)\log n)$。

 

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int N=100010,M=N*2;
const double inf=1e20;
const double eps=1e-9;
int n,m,i,x,y,val[N],ce,cb,q[N][3],ans;double X;
struct Shape{
  bool type;
  double x,y,r,a[35][2];
  int n;
  void read(){
    char t[5];
    scanf("%s",t);
    if(t[0]=='C')type=0,scanf("%lf%lf%lf",&x,&y,&r);
    else{
      type=1;
      scanf("%d",&n);
      for(int i=0;i<n;i++)scanf("%lf%lf",&a[i][0],&a[i][1]);
      a[n][0]=a[0][0],a[n][1]=a[0][1];
    }
  }
  double getl(){
    if(!type)return x-r;
    double ret=inf;
    for(int i=0;i<n;i++)ret=min(ret,a[i][0]);
    return ret;
  }
  double getr(){
    if(!type)return x+r;
    double ret=-inf;
    for(int i=0;i<n;i++)ret=max(ret,a[i][0]);
    return ret;
  }
  double getd(double o){
    if(!type)return y-sqrt(max(r*r-(o-x)*(o-x),0.0));
    double ret=inf;
    for(int i=0;i<n;i++){
      double A=a[i][0],B=a[i][1],C=a[i+1][0],D=a[i+1][1];
      if(A>C)swap(A,C),swap(B,D);
      if(o<A-eps||o>C+eps)continue;
      if(o<A+eps){
        ret=min(ret,B);
        continue;
      }
      if(o>C-eps){
        ret=min(ret,D);
        continue;
      }
      ret=min(ret,B+(D-B)/(C-A)*(o-A));
    }
    return ret;
  }
  double getu(double o){
    if(!type)return y+sqrt(max(r*r-(o-x)*(o-x),0.0));
    double ret=-inf;
    for(int i=0;i<n;i++){
      double A=a[i][0],B=a[i][1],C=a[i+1][0],D=a[i+1][1];
      if(A>C)swap(A,C),swap(B,D);
      if(o<A-eps||o>C+eps)continue;
      if(o<A+eps){
        ret=max(ret,B);
        continue;
      }
      if(o>C-eps){
        ret=max(ret,D);
        continue;
      }
      ret=max(ret,B+(D-B)/(C-A)*(o-A));
    }
    return ret;
  }
}a[N];
struct E{
  double x;int y,t;
  E(){}
  E(double _x,int _y,int _t){x=_x,y=_y,t=_t;}
}e[N*4];
inline bool cmp(const E&a,const E&b){
  if(a.t<2&&b.t<2&&a.y==b.y)return a.t<b.t;
  return a.x<b.x;
}
struct P{double x,y;}b[M];
int from[M],son[M][2],f[M],root,L,R,fa[N],g[N],v[M],nxt[M],ed,st[N],en[N],dfn,bit[M];
inline void rotate(int x){
  int y=f[x],w=son[y][1]==x;
  son[y][w]=son[x][w^1];
  if(son[x][w^1])f[son[x][w^1]]=y;
  if(f[y]){
    int z=f[y];
    if(son[z][0]==y)son[z][0]=x;
    if(son[z][1]==y)son[z][1]=x;
  }
  f[x]=f[y];son[x][w^1]=y;f[y]=x;
}
inline void splay(int x,int w){
  while(f[x]!=w){
    int y=f[x];
    if(f[y]!=w){if((son[f[y]][0]==y)^(son[y][0]==x))rotate(x);else rotate(y);}
    rotate(x);
  }
  if(!w)root=x;
}
inline bool bigger(int x,int y){
  if(y==L)return 1;
  if(y==R)return 0;
  if(x+n==y||x==y+n)return x>y;
  double A=x<=n?a[x].getd(X):a[x-n].getu(X),
         B=y<=n?a[y].getd(X):a[y-n].getu(X);
  return A>B;
}
inline bool biggerp(double x,int y){
  if(y==L)return 1;
  if(y==R)return 0;
  double B=y<=n?a[y].getd(X):a[y-n].getu(X);
  return x>B;
}
void ins(int x,int y){
  int w=bigger(y,x);
  if(!son[x][w]){son[x][w]=y;f[y]=x;return;}
  ins(son[x][w],y);
}
int ask(int x,double y){
  if(!x)return 0;
  if(biggerp(y,x)){
    int t=ask(son[x][1],y);
    return t?t:x;
  }
  return ask(son[x][0],y);
}
inline void add(int x){
  ins(root,x);
  ins(root,x+n);
  splay(x,0);
  int y;
  for(y=son[x][0];son[y][1];y=son[y][1]);
  if(y<L)fa[x]=y<=n?y:fa[y-n];
  splay(y,0);
}
inline void del(int x){
  splay(x,0);
  int y;
  for(y=son[x][0];son[y][1];y=son[y][1]);
  splay(y,x);
  son[y][1]=son[x][1];
  f[son[x][1]]=y;
  f[root=y]=0;
}
inline void getpos(int x){
  int y=ask(root,b[x].y);
  if(y<L)from[x]=y<=n?y:fa[y-n];
  splay(y,0);
}
void dfs(int x){
  st[x]=++dfn;
  for(int i=g[x];i;i=nxt[i])dfs(i);
  en[x]=++dfn;
}
inline void modify(int x,int y){for(;x<=dfn;x+=x&-x)bit[x]^=y;}
inline int sum(int x){int t=0;for(;x;x-=x&-x)t^=bit[x];return t;}
int main(){
  scanf("%d%d",&n,&m);
  for(i=1;i<=n;i++){
    a[i].read();
    scanf("%d",&val[i]);
    e[++ce]=E(a[i].getl(),i,0);
    e[++ce]=E(a[i].getr(),i,1);
  }
  for(i=1;i<=m;i++){
    char op[5];
    scanf("%s",op);
    if(op[0]=='Q'){
      cb++;
      scanf("%lf%lf",&b[cb].x,&b[cb].y);
      e[++ce]=E(b[cb].x,cb,2);
      cb++;
      scanf("%lf%lf",&b[cb].x,&b[cb].y);
      e[++ce]=E(b[cb].x,cb,2);
      q[i][1]=cb-1,q[i][2]=cb;
    }else{
      q[i][0]=1;
      scanf("%d%d",&q[i][1],&q[i][2]);
    }
  }
  sort(e+1,e+ce+1,cmp);
  L=n*2+1;R=L+1;
  son[L][1]=R,f[R]=L;
  root=L;
  for(i=1;i<=ce;i++){
    X=e[i].x;
    if(!e[i].t)add(e[i].y);
    else if(e[i].t==1)del(e[i].y),del(e[i].y+n);
    else getpos(e[i].y);
  }
  for(i=1;i<=n;i++)if(fa[i])nxt[i]=g[fa[i]],g[fa[i]]=i;
  for(i=1;i<=n;i++)if(!fa[i])dfs(i);
  for(i=1;i<=n;i++)modify(st[i],val[i]),modify(en[i],val[i]);
  for(i=1;i<=m;i++)if(q[i][0]){
    x=q[i][1],y=q[i][2]^val[x];
    val[x]=q[i][2];
    modify(st[x],y),modify(en[x],y);
  }else printf("%d\n",ans^=sum(st[from[q[i][1]]])^sum(st[from[q[i][2]]]));
  return 0;
}