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題面

題解

不難看出每個點的大小為行列限制中較小的那一個（因為資料保證有解）
對於行的每個限制，能取到的個數是列裡限制大於等於它的數的個數，同理，對於列是行裡大於它的個數（這裡沒有等於，為了避免重複計算）
於是可以對於行列分別開權值線段樹，修改的時候只要把對應的貢獻改一下就好了

//minamoto
#include<bits/stdc++.h>
#define R register
#define ll long long
#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;}
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 C=-1,Z=0;
inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}
void print(R ll x){
    if(C>1<<20)Ot();if(x<0)sr[++C]='-',x=-x;
    while(z[++Z]=x%10+48,x/=10);
    while(sr[++C]=z[Z],--Z);sr[++C]='\n';
}
const int N=1e5+5;
struct change{int op,pos,x;}c[N];
ll res;int n,m,a[N],b[N],aa[N],bb[N],op,pos,x,lim;
struct seg{
    struct node{int ls,rs,cnt;ll sum;}t[N<<5];
    int rt,tot;
    void ins(int &p,int l,int r,int x,int ty){
        if(!p)p=++tot;t[p].cnt+=ty,t[p].sum+=x*ty;
        if(l==r)return;int mid=(l+r)>>1;
        x<=mid?ins(t[p].ls,l,mid,x,ty):ins(t[p].rs,mid+1,r,x,ty);
    }
    int q_cnt(int p,int l,int r,int ql,int qr){
        if(!p)return 0;if(ql<=l&&qr>=r)return t[p].cnt;
        int mid=(l+r)>>1,res=0;
        if(ql<=mid)res+=q_cnt(t[p].ls,l,mid,ql,qr);
        if(qr>mid)res+=q_cnt(t[p].rs,mid+1,r,ql,qr);
        return res;
    }
    ll q_sum(int p,int l,int r,int ql,int qr){
        if(!p)return 0;if(ql<=l&&qr>=r)return t[p].sum;
        int mid=(l+r)>>1;ll res=0;
        if(ql<=mid)res+=q_sum(t[p].ls,l,mid,ql,qr);
        if(qr>mid)res+=q_sum(t[p].rs,mid+1,r,ql,qr);
        return res;
    }
}A,B;
int main(){
    freopen("graph.in","r",stdin);
    freopen("graph.out","w",stdout);
    n=read();
    fp(i,1,n)aa[i]=a[i]=read(),cmax(lim,a[i]);
    fp(i,1,n)bb[i]=b[i]=read(),cmax(lim,b[i]);
    m=read();
    fp(i,1,m)c[i].op=read(),c[i].pos=read(),c[i].x=read(),cmax(lim,c[i].x);
    sort(aa+1,aa+1+n),sort(bb+1,bb+1+n);
    for(R int i=1,j=1;i<=n;++i){
        while(j<=n&&aa[i]>bb[j])++j;
        res+=1ll*aa[i]*(n-j+1);
    }for(R int i=1,j=1;i<=n;++i){
        while(j<=n&&bb[i]>=aa[j])++j;
        res+=1ll*bb[i]*(n-j+1);
    }print(res);
    fp(i,1,n)A.ins(A.rt,0,lim,a[i],1);
    fp(i,1,n)B.ins(B.rt,0,lim,b[i],1);
    fp(i,1,m){
        op=c[i].op,pos=c[i].pos,x=c[i].x;
        if(op==0){
            res-=1ll*B.q_cnt(B.rt,0,lim,a[pos],lim)*a[pos];
            if(a[pos]>0)res-=B.q_sum(B.rt,0,lim,0,a[pos]-1);
            A.ins(A.rt,0,lim,a[pos],-1);
            a[pos]=x;
            A.ins(A.rt,0,lim,a[pos],1);
            res+=1ll*B.q_cnt(B.rt,0,lim,a[pos],lim)*a[pos];
            if(a[pos]>0)res+=B.q_sum(B.rt,0,lim,0,a[pos]-1);
        }else{
            if(b[pos]<lim)res-=1ll*A.q_cnt(A.rt,0,lim,b[pos]+1,lim)*b[pos];
            res-=A.q_sum(A.rt,0,lim,0,b[pos]);
            B.ins(B.rt,0,lim,b[pos],-1);
            b[pos]=x;
            B.ins(B.rt,0,lim,b[pos],1);
            if(b[pos]<lim)res+=1ll*A.q_cnt(A.rt,0,lim,b[pos]+1,lim)*b[pos];
            res+=A.q_sum(A.rt,0,lim,0,b[pos]);
        }print(res);
    }return Ot(),0;
}