luoguP3415 祭坛

https://www.luogu.org/problemnew/show/P3415

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 <bits/stdc++.h>

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;

}