考虑二分结界层数，将 n 个点按 x 大小依次加入答案，一行一行的做，用树状数组维护当前这一行中[0, x - 1] 包含祭坛大于 mid 的且 [x + 1, n] 中包含的祭坛也大于 mid 的坐标，再计算出这一行有几个地方可以作为中心，简单容斥（可能还不算容斥？）一下就可以了

有一个坑就是当节点层数是 0 的时候输出两行 0

#include 
    
     using namespace std;const int N = 1e5 + 5;struct ele {    int x, y;    bool operator < (const ele A) const {return x < A.x || (x == A.x && y < A.y);}}d[N];int sum[N], f[N], y[N], allsum[N];int n, len;inline int lowbit(int x) {return x & -x;}inline void add(int x, int y) {for(int i = x; i <= n; i += lowbit(i)) f[i] += y;}inline int query(int x) {int ans = 0; for(int i = x; i; i -= lowbit(i)) ans += f[i]; return ans;}int solve(int mid) {    memset(sum, 0, sizeof(sum));    memset(f, 0, sizeof(f));    int now = 1, ans = 0;    for(int i = 0; i <= n; i++) {        len = 0;        while(now <= n && d[now].x == i) {            int t = d[now++].y;            y[++len] = t;        }        if(len >= (mid << 1)) {            int l = y[mid] + 1, r = y[len - mid + 1] - 1;            for(int j = mid + 1; j <= len - mid; j++) ans -= (sum[y[j]] >= mid && (allsum[y[j]] - sum[y[j]] >= mid));            ans += (query(r) - query(l - 1));        }        for(int j = 1; j <= len; j++) {            int t = y[j];            if(sum[t] >= mid && (allsum[t] - sum[t] == mid)) add(t, -1);            sum[t]++;            if(sum[t] == mid && (allsum[t] - sum[t] >= mid)) add(t, 1);        }    }    return ans;}int main() {    cin >> n;    for(int i = 1; i <= n; i++) scanf("%d %d", &d[i].x, &d[i].y);    sort(d + 1, d + n + 1);    for(int i = 1; i <= n; i++) allsum[d[i].y]++;    int l = 1, r = n / 4;    while(l < r) {        int mid = (l + r + 1) >> 1;        if(solve(mid)) l = mid;        else r = mid - 1;    }    if(solve(l) == 0) printf("0\n0\n");    else printf("%d\n%d\n", l, solve(l));    return 0;}