题面

传送门

题解

首先\(x\)和\(y\)两维互相独立，可以分开考虑，我们以\(x\)为例

我们把\(x\)做个前缀和，那么就是问有多少\(i\)满足\(s_is_{i-1}<0\)，其中\(s_0=1\)。这个条件等价于\(\max(s_i,s_{i-1})>0\)且\(\min(s_i,s_{i-1})<0\)。我们可以容斥一下，就是总数减去\(\max(s_i,s_{i-1})<0\)的个数减去\(\min(s_i,s_{i-1})>0\)的个数

注意到一次单点修改会使一个点到结尾的\(s_i\)区间加，而对应的\(\max(s_i,s_{i+1}),\max(s_{i+1},s_{i+2})\)之类的可以直接加上这个值，但前面的不行。我们记录一个偏移量，并且开一个平衡树，平衡树里只放指针右边的元素，那么一次修改不会影响前面，我们可以在移动指针的时候顺便维护答案，而指针右边的答案在平衡树里找就可以了

//minamoto
#include<bits/stdc++.h>
#define R register
#define inline __inline__ __attribute__((always_inline))
#define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)
#define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
template<class T>inline bool cmax(T&a,const T&b){return a<b?a=b,1:0;}
template<class T>inline bool cmin(T&a,const T&b){return a>b?a=b,1:0;}
using namespace std;
char buf[1<<21],*p1=buf,*p2=buf;
inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}
int read(){
    R int res,f=1;R char ch;
    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
    for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');
    return res*f;
}
char sr[1<<21],z[20];int K=-1,Z=0;
inline void Ot(){fwrite(sr,1,K+1,stdout),K=-1;}
void print(R int x){
    if(K>1<<20)Ot();if(x<0)sr[++K]='-',x=-x;
    while(z[++Z]=x%10+48,x/=10);
    while(sr[++K]=z[Z],--Z);sr[++K]='\n';
}
inline char getop(){char ch;while((ch=getc())>'Z'||ch<'A');return ch;}
unsigned int aaa=19260817;
inline unsigned int rd(){aaa^=aaa>>15,aaa+=aaa<<12,aaa^=aaa>>3;return aaa;}
inline int max(R int x,R int y){return x>y?x:y;}
inline int min(R int x,R int y){return x<y?x:y;}
const int N=2e5+5;
struct node;typedef node* ptr;
struct node{
    ptr lc,rc;int sz,v;unsigned int pr;
    inline ptr init(R int val){return sz=1,v=val,pr=rd(),this;}
    inline ptr upd(){return sz=lc->sz+rc->sz+1,this;}
};
struct Treap{
    node e[N],*rt=e,*pp=e;
    inline ptr newnode(R int v){return ++pp,pp->lc=pp->rc=e,pp->init(v);}
    void split(ptr p,int k,ptr &s,ptr &t){
        if(p==e)return s=t=e,void();
        if(p->v<=k)s=p,split(p->rc,k,p->rc,t);
            else t=p,split(p->lc,k,s,p->lc);
        p->upd();
    }
    ptr merge(ptr s,ptr t){
        if(s==e)return t;if(t==e)return s;
        if(s->pr<t->pr)return s->rc=merge(s->rc,t),s->upd();
        return t->lc=merge(s,t->lc),t->upd();
    }
    void push(int k){
        ptr s,t;
        split(rt,k,s,t),rt=merge(merge(s,newnode(k)),t);
    }
    void pop(int k){
        ptr s,t,p,q;
        split(rt,k,s,t),split(s,k-1,p,q),q=merge(q->lc,q->rc);
        rt=merge(merge(p,q),t);
    }
    int les(int k){
        ptr s,t;int now;
        split(rt,k-1,s,t),now=s->sz;
        return rt=merge(s,t),now;
    }
    int gre(int k){
        ptr s,t;int now;
        split(rt,k,s,t),now=t->sz;
        return rt=merge(s,t),now;
    }
    Treap(){e->lc=e->rc=e;}
}sx[2],sy[2];
int nx[N],ny[N],X,Y,dx,dy,p,ansx,ansy,n;
void Ri(){
    if(p==n)return;++p;
    sx[0].pop(min(X,X+nx[p])-dx),sx[1].pop(max(X,X+nx[p])-dx),
    sy[0].pop(min(Y,Y+ny[p])-dy),sy[1].pop(max(Y,Y+ny[p])-dy);
    if(X*(X+nx[p])<0)++ansx;if(Y*(Y+ny[p])<0)++ansy;
    X+=nx[p],Y+=ny[p];
}
void Le(){
    if(p==1)return;X-=nx[p],Y-=ny[p];
    if(X*(X+nx[p])<0)--ansx;if(Y*(Y+ny[p])<0)--ansy;
    sx[0].push(min(X,X+nx[p])-dx),sx[1].push(max(X,X+nx[p])-dx),
    sy[0].push(min(Y,Y+ny[p])-dy),sy[1].push(max(Y,Y+ny[p])-dy);
    --p;
}
void modify(){
    int x=read(),y=read();
    if(X*(X-nx[p])<0)--ansx;if(Y*(Y-ny[p])<0)--ansy;
    X+=x-nx[p],Y+=y-ny[p],dx+=x-nx[p],dy+=y-ny[p],nx[p]=x,ny[p]=y;
    if(X*(X-x)<0)++ansx;if(Y*(Y-y)<0)++ansy;
}
inline int query(){return n-p-sx[0].gre(-dx)-sx[1].les(-dx)+n-p-sy[0].gre(-dy)-sy[1].les(-dy)+ansx+ansy;}
int main(){
//  freopen("testdata.in","r",stdin);
    n=read(),X=Y=1;
    fp(i,1,n){
        nx[i]=read(),ny[i]=read();
        sx[0].push(min(X,X+nx[i])),sx[1].push(max(X,X+nx[i]));
        sy[0].push(min(Y,Y+ny[i])),sy[1].push(max(Y,Y+ny[i]));
        X+=nx[i],Y+=ny[i];
    }
    X=Y=1,p=0,Ri();
    for(int q=read();q;--q){
        char op=getop();
        switch(op){
            case 'F':Ri();break;
            case 'B':Le();break;
            case 'C':modify();break;
            case 'Q':print(query());break;
        }
    }
    return Ot(),0;
}