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;
}

原文地址：https://www.cnblogs.com/LJC00118/p/9656893.html