[Source unknown] Vulcan fish

The meaning of problems

　　There are two-dimensional plane \ (n-\) points, there is a net four sides parallel to the coordinate axes, while the plane \ (n-\) numbered as \ (1 \) to the \ (n-\) point.
　　You need to support \ (3 \) modes of operation:
　　　　1 lrd: number in \ ([l, r] \ ) at all points to within \ (X \) translational axis positive direction \ (D \) units
　　　　1 lrd: No in \ ([l, r] \ ) at all points to within \ (Y \) axis positive direction translation \ (D \) units
　　　　3 lr: seeking the number of points in the nets (points on nets boundary can be considered ) \
　　(T \ in [1,10] \) set of data, \ (. 1 \ n-Le, m \ 30000 Le, \ Space. 1 \ Le D \ ^. 9 Le 10, \ Space x_1 \ Le x_2, \ Space Y_1 \ le y_2 \)

answer

　　Universal problem, the net was split into \ (4 \) a lower left corner of the rectangular infinity, each sub-question becomes limited only points \ (x, y \) coordinate of the upper limit, before each operation of the cross-border points deleted i.e. can, which only supports single tree line and seeking to delete the maximum range. Finally, inclusion and exclusion at \ (4 \) rectangular answers.

　　Or just able to save the title.

#include<bits/stdc++.h>
#define ll long long
#define N 30005
#define lowbit(x) x&-x
const ll inf = 1e9 + 7;
using namespace std;
inline int read(){
    int x=0; bool f=1; char c=getchar();
    for(;!isdigit(c); c=getchar()) if(c=='-') f=0;
    for(; isdigit(c); c=getchar()) x=(x<<3)+(x<<1)+(c^'0');
    if(f) return x; return 0-x;
}
 
int n,m,x1,Y1,x2,y2,p[N][2],ans[N];
struct query{int opt,l,r,d;}q[N];
struct SegTree{
    #define ls o<<1
    #define rs o<<1|1
    ll mx[N<<2],tag[N<<2];
    inline void pushdown(int o){
        if(tag[o]){
            mx[ls]+=tag[o], mx[rs]+=tag[o];
            tag[ls]+=tag[o], tag[rs]+=tag[o];
            tag[o]=0;
        }
    }
    inline void pushup(int o){
        mx[o]=max(mx[ls],mx[rs]);
    }
    void build(int o, int l, int r, int opt){
        tag[o]=0;
        if(l==r){mx[o]=p[l][opt]; return;}
        int mid=(l+r)>>1;
        build(ls,l,mid,opt);
        build(rs,mid+1,r,opt);
        pushup(o);
    }
    void mdf(int o, int l, int r, int L, int R, ll v){
        if(L<=l && r<=R){
            mx[o]+=v, tag[o]+=v;
            return;
        }
        pushdown(o);
        int mid=(l+r)>>1;
        if(L<=mid) mdf(ls,l,mid,L,R,v);
        if(R>mid) mdf(rs,mid+1,r,L,R,v);
        pushup(o);
    }
    int query(int o, int l, int r){
        if(l==r) return l;
        pushdown(o);
        int mid=(l+r)>>1;
        return mx[ls]>mx[rs] ? query(ls,l,mid) : query(rs,mid+1,r);
    }
}tx,ty;
 
int fwk[N];
void ins(int x,int v){for(; x<=n; x+=lowbit(x)) fwk[x]+=v;}
int ask(int x){
    int res=0;
    for(; x; x-=lowbit(x)) res+=fwk[x];
    return res;
}
 
void calc(int x, int y, int v){
    tx.build(1,1,n,0);
    ty.build(1,1,n,1);
    memset(fwk,0,sizeof(fwk));
    for(int i=1; i<=n; i++) ins(i,1);
    for(int i=1; i<=m; i++){
        while(tx.mx[1]>x){
            int p=tx.query(1,1,n);
            tx.mdf(1,1,n,p,p,-inf);
            if(ask(p)-ask(p-1)) ins(p,-1);
        }
        while(ty.mx[1]>y){
            int p=ty.query(1,1,n);
            ty.mdf(1,1,n,p,p,-inf);
            if(ask(p)-ask(p-1)) ins(p,-1);
        }
        if(q[i].opt==1) tx.mdf(1,1,n,q[i].l,q[i].r,(ll)q[i].d);
        if(q[i].opt==2) ty.mdf(1,1,n,q[i].l,q[i].r,(ll)q[i].d);
        if(q[i].opt==3) ans[i] += v*(ask(q[i].r)-ask(q[i].l-1));
    }
}
signed main(){
    //freopen("c.in","r",stdin);
    //freopen("std.out","w",stdout);
    int T=read();
    while(T--){
        n=read();
        x1=read(), Y1=read(), x2=read(), y2=read();
        for(int i=1; i<=n; i++) p[i][0]=read(), p[i][1]=read();
        m=read();
        for(int i=1; i<=m; i++){
            q[i].opt=read(), q[i].l=read(), q[i].r=read();
            if(q[i].opt!=3) q[i].d=read();
            ans[i]=0;
        }
        calc(x2,y2,1);
        calc(x1-1,Y1-1,1), calc(x2,Y1-1,-1), calc(x1-1,y2,-1);
        for(int i=1;i<=m;i++) if(q[i].opt==3) printf("%d\n",ans[i]);
    }
    return 0;
}