题目链接：https://atcoder.jp/contests/abc131/tasks/abc131_f

转自博客：https://blog.csdn.net/qq_37656398/article/details/93496476

题目大意

　　给定 N 个点，若有如下图所示的三个点（黑点）就可以构造岀一个新的点（红点）。求可构造的点的最多的个数。

分析

　　首先，如果两个点的某一个坐标相等，我们定义这两个点是联通的。

　　其次，如果红点能够被构造，它必然是由处于同一连通块中的点所构造的。

　　于是，只要分别计算每个联通块能构造多少点就可以了。

　　对于每个连通块，把 x 坐标和 y 坐标单独弄一个集合，然后就在两个集合中不断分别取两个进行构造，直到不能构造为止，图示如下：

　　我们发现，构造的过程就是把构造前的二部图构造成完全二部图，每多一条边就表明构造了一个点。

　　于是把所有连通块所对应的完全二部图边数都加起来再减去 N 就能得到能构造的点的数目了。

代码如下

#include
using namespace std;

#define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define Rep(i,n) for (int i = 0; i < (n); ++i) #define For(i,s,t) for (int i = (s); i <= (t); ++i) #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
#define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i) #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)

#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl

#define LOWBIT(x) ((x)&(-x))

#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define UNIQUE(x) x.erase(unique(x.begin(), x.end()), x.end())
#define REMOVE(x, c) x.erase(remove(x.begin(), x.end(), c), x.end()); // 删去 x 中所有 c
#define TOLOWER(x) transform(x.begin(), x.end(), x.begin(),::tolower);
#define TOUPPER(x) transform(x.begin(), x.end(), x.begin(),::toupper);

#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,inf,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))

#define MP make_pair
#define PB push_back
#define ft first
#define sd second

template
istream &operator>>(istream &in, pair &p) {
in >> p.first >> p.second;
return in;
}

template
istream &operator>>(istream &in, vector &v) {
for (auto &x: v)
in >> x;
return in;
}

template
ostream &operator<<(ostream &out, const std::pair &p) {
out << "[" << p.first << ", " << p.second << "]" << "\n";
return out;
}

inline int gc(){
static const int BUF = 1e7;
static char buf[BUF], *bg = buf + BUF, *ed = bg;

 if(bg == ed) fread(bg = buf, , BUF, stdin);  
 return \*bg++;  

}

inline int ri(){
int x = , f = , c = gc();
for(; c<||c>; f = c=='-'?-:f, c=gc());
for(; c>&&c<; x = x* + c - , c=gc());
return x*f;
}

template
inline string toString(T x) {
ostringstream sout;
sout << x;
return sout.str();
}

inline int toInt(string s) {
int v;
istringstream sin(s);
sin >> v;
return v;
}

//min <= aim <= max template
inline bool BETWEEN(const T aim, const T min, const T max) {
return min <= aim && aim <= max;
}

typedef long long LL;
typedef unsigned long long uLL;
typedef pair< double, double > PDD;
typedef pair< int, int > PII;
typedef pair< int, PII > PIPII;
typedef pair< string, int > PSI;
typedef pair< int, PSI > PIPSI;
typedef set< int > SI;
typedef set< PII > SPII;
typedef vector< int > VI;
typedef vector< double > VD;
typedef vector< VI > VVI;
typedef vector< SI > VSI;
typedef vector< PII > VPII;
typedef map< int, int > MII;
typedef map< int, string > MIS;
typedef map< int, PII > MIPII;
typedef map< PII, int > MPIII;
typedef map< string, int > MSI;
typedef map< string, string > MSS;
typedef map< PII, string > MPIIS;
typedef map< PII, PII > MPIIPII;
typedef multimap< int, int > MMII;
typedef multimap< string, int > MMSI;
//typedef unordered_map< int, int > uMII;
typedef pair< LL, LL > PLL;
typedef vector< LL > VL;
typedef vector< VL > VVL;
typedef priority_queue< int > PQIMax;
typedef priority_queue< int, VI, greater< int > > PQIMin;
const double EPS = 1e-;
const LL inf = 0x7fffffff;
const LL infLL = 0x7fffffffffffffffLL;
const LL mod = 1e9 + ;
const int maxN = 1e5 + ;
const LL ONE = ;
const LL evenBits = 0xaaaaaaaaaaaaaaaa;
const LL oddBits = 0x5555555555555555;

int N, K, M;
LL ans;

int f[maxN << ], sz[maxN << ][];

int Find(int x){
while (x != f[x]) x = f[x] = f[f[x]];
return x;
}

int main(){
//freopen("MyOutput.txt","w",stdout);
//freopen("input.txt","r",stdin);
//INIT();
cin >> N;
Rep(i, maxN << ) f[i] = i;

 Rep(i, N) {  
     int x, y;  
     cin >> x >> y;  
     y += maxN;  
     f\[Find(x)\] = Find(y);  
 }  
 Rep(i, maxN << ) ++sz\[Find(i)\]\[i / maxN\];  
 Rep(i, maxN << ) ans += (LL)sz\[i\]\[\] \* sz\[i\]\[\];  
 cout << ans - N << endl;  
 return ;  

}