Effect: to the point in planes $ $ n, define the set $ S (l, r, a) $ abscissa represents all points $ [l, r] $ ordinate $ [a, \ infty] $ within the request can. how many different collections available.
From the enumeration bottom right-most point, stats Contribution Fenwick tree down.
#include <iostream>
#include <algorithm>
#include <cstdio>
#define REP(i,a,n) for(int i=a;i<=n;++i)
using namespace std;
typedef long long ll;
const int N = 1e6+10, INF = 0x3f3f3f3f;
int n, c[N], vis[N], b[N];
struct _ {
int x,y;
bool operator < (const _&rhs) const {
if (y!=rhs.y) return y>rhs.y;
return x<rhs.x;
}
} a[N];
int ID(int x) {
return lower_bound(b+1,b+1+*b,x)-b;
}
int qry(int x) {
int r = 0;
for (; x; x^=x&-x) r+=c[x];
return r;
}
int qry(int l, int r) {
return qry(ID(r))-qry(ID(l-1));
}
void add(int x) {
x = ID(x);
if (!vis[x]) {
vis[x] = 1;
for (; x<=*b; x+=x&-x) ++c[x];
}
}
int main() {
scanf("%d", &n);
REP(i,1,n) {
scanf("%d%d",&a[i].x,&a[i].y);
b[++*b]=a[i].x;
b[++*b]=a[i].x-1;
}
b[++*b]=0,b[++*b]=INF;
sort(b+1,b+1+*b),*b=unique(b+1,b+1+*b)-b-1;
sort(a+1,a+1+n);
ll ans = 0;
REP(i,1,n) {
int j=i;
while (j<n&&a[j+1].y==a[i].y) ++j;
REP(k,i,j) {
int L = qry(1,a[k].x-1);
int R = qry(a[k].x+1,k==j?INF:a[k+1].x-1);
ans += (ll)(L+1)*(R+1);
add(a[k].x);
}
i = j;
}
printf("%lld\n", ans);
}