CF1140F
给定一个点集，初始为空。
q次操作，每次操作给定一个点，若点集中存在这个点就删除这个点，否则插入这个点。
每次操作完之后，输出反复执行如下操作后点集的大小：
找出x1,x2,y1,y2使得(x1,y1)(x1,y2)(x2,y1)都在点集中，且(x2,y2)不在，并将(x2,y2)加入点集。
\(q\le 3\cdot 10^5,TL=3.5s,ML=1G\)
题解
考虑每个点相当于二分图中的一条边，那么答案就是二分图每个联通块中x方点和y方点点数乘积之和。
删除看上去很麻烦，可以考虑离线之后变成对区间内的询问。
对询问建一棵线段树，最后dfs的时候，每遍历到一个节点，先插入并记录修改的部分，然后如果是叶子就输出答案，否则递归，最后撤回修改。
/*
Author: QAQ-Automaton
LANG: C++
PROG: F.cpp
Mail: cnyalilk@vip.qq.com
*/
#include<bits/stdc++.h>
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define DEBUG printf("Passing [%s] in LINE %lld\n",__FUNCTION__,__LINE__)
#define Debug debug("Passing [%s] in LINE %lld\n",__FUNCTION__,__LINE__)
#define all(x) x.begin(),x.end()
#define x first
#define y second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pii;
#define inf 0x3f3f3f3f3f3f3f3f3f3f3f3f3f3f3f3f
const double eps=1e-8;
const double pi=acos(-1.0);
template<class T>ll chkmin(T &a,T b){return a>b?a=b,1:0;}
template<class T>ll chkmax(T &a,T b){return a<b?a=b,1:0;}
template<class T>T sqr(T a){return a*a;}
template<class T>T mmin(T a,T b){return a<b?a:b;}
template<class T>T mmax(T a,T b){return a>b?a:b;}
template<class T>T aabs(T a){return a<0?-a:a;}
template<class T>ll dcmp(T a,T b){return a>b;}
template<ll *a>ll cmp_a(ll x,ll y){return a[x]<a[y];}
#define min mmin
#define max mmax
#define abs aabs
namespace io {
const ll SIZE = (1 << 21) + 1;
char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55]; ll f, qr;
// getchar
#define gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
// prll the remaining part
inline void flush () {
fwrite (obuf, 1, oS - obuf, stdout);
oS = obuf;
}
// putchar
inline void putc (char x) {
*oS ++ = x;
if (oS == oT) flush ();
}
// input a signed lleger
inline bool read (signed &x) {
for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
return 1;
}
inline bool read (long long &x) {
for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
return 1;
}
inline bool read (char &x) {
x=gc();
return x!=EOF;
}
inline bool read(char *x){
while((*x=gc())=='\n' || *x==' '||*x=='\r')if(*x==EOF)return 0;
while(!(*x=='\n'||*x==' '||*x=='\r'))*(++x)=gc();
*x=0;
return 1;
}
template<typename A,typename ...B>
inline bool read(A &x,B &...y){
return read(x)&&read(y...);
}
// prll a signed lleger
inline bool write (signed x) {
if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
while (x) qu[++ qr] = x % 10 + '0', x /= 10;
while (qr) putc (qu[qr --]);
return 0;
}
inline bool write (long long x) {
if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
while (x) qu[++ qr] = x % 10 + '0', x /= 10;
while (qr) putc (qu[qr --]);
return 0;
}
inline bool write (char x) {
putc(x);
return 0;
}
inline bool write(const char *x){
while(*x){putc(*x);++x;}
return 0;
}
inline bool write(char *x){
while(*x){putc(*x);++x;}
return 0;
}
template<typename A,typename ...B>
inline bool write(A x,B ...y){
return write(x)||write(y...);
}
//no need to call flush at the end manually!
struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using io :: read;
using io :: putc;
using io :: write;
map<pii,ll> qwq;
ll fa[600005],cnt0[600005],cnt1[600005],ans;
ll a[600005],b[600005],c[600005],d[600005],t;
ll find(ll x){return x==fa[x]?x:find(fa[x]);}
struct smt{
ll ls,rs;
vector<pii> addlist;
smt *l,*r;
smt(ll la,ll ra){
ls=la;rs=ra;
addlist.clear();
if(ls==rs){l=r=0;return;}
ll mid=(ls+rs)>>1;
l=new smt(ls,mid);
r=new smt(mid+1,rs);
}
void add(ll la,ll ra,pii w){
if(la<=ls && rs<=ra){addlist.push_back(w);return;}
if(la<=l->rs)l->add(la,ra,w);
if(r->ls<=ra)r->add(la,ra,w);
}
void calc(){
ll nt=t,oa=ans;
for(auto i:addlist){
if(find(i.x)!=find(i.y+300000)){
ll u=find(i.x),v=find(i.y+300000);
if(cnt0[u]+cnt1[u]>cnt0[v]+cnt1[v])swap(u,v);
++t;
c[t]=cnt0[v];d[t]=cnt1[v];
b[t]=v;
a[t]=u;
fa[u]=v;
ans-=cnt0[u]*cnt1[u]+cnt0[v]*cnt1[v];
cnt0[v]+=cnt0[u];
cnt1[v]+=cnt1[u];
ans+=cnt0[v]*cnt1[v];
}
}
if(ls==rs){write(ans,' ');}
else{
l->calc();
r->calc();
}
while(t>nt){
fa[a[t]]=a[t];
cnt0[b[t]]=c[t];
cnt1[b[t]]=d[t];
--t;
}
ans=oa;
}
};
smt *rt;
int main(){
#ifdef QAQAutoMaton
freopen("F.in","r",stdin);
freopen("F.out","w",stdout);
#endif
for(ll i=1;i<=600000;++i){fa[i]=i;cnt0[i]=i<=300000;cnt1[i]=!cnt0[i];}
ll q;
pii a;
read(q);
rt=new smt(1,q);
for(ll i=1;i<=q;++i){
read(a.x,a.y);
if(qwq.count(a)){
rt->add(qwq[a],i-1,a);
qwq.erase(a);
}
else qwq[a]=i;
}
for(auto i:qwq){
rt->add(i.y,q,i.x);
}
rt->calc();
return 0;
}