（ 2 ） （矩形计数） 平面上有 $n$ 个关键点，求有多少个四条边都和 $x$ 轴或者 $y$ 轴平行的矩形，满足四个顶点都是关键点。给出的关键点可能有重复，但完全重合的矩形只计一 次。

试补全枚举算法。

1 2#include <iostream>3using namespace std;4 5struct point {6    int x, y, id;7};8 9bool equals(point a, point b) {10    return a.x == b.x && a.y == b.y;11}12 13bool cmp(point a, point b) {14    return ①;15}16 17void sort(point A[], int n) {18    for (int i = 0; i < n; i++)19        for (int j = 1; j < n; j++)20            if (cmp(A[j], A[j - 1])) {21                point t = A[j];22                A[j] = A[j - 1];23                A[j - 1] = t;24            }25}26 27int unique(point A[], int n) {28    int t = 0;29    for (int i = 0; i < n; i++)30        if (②)31            A[t++] = A[i];32    return t;33}34 35bool binary_search(point A[], int n, int x, int y) {36    point p;37    p.x = x;38    p.y = y;39    p.id = n;40    int a = 0, b = n - 1;41    while (a < b) {42        int mid = ③;43        if (④)44            a = mid + 1;45        else46            b = mid;47    }48    return equals(A[a], p);49}50 51const int MAXN = 1000;52point A[MAXN];53 54int main() {55    int n;56    cin >> n;57    for (int i = 0; i < n; i++) {58        cin >> A[i].x >> A[i].y;59        A[i].id = i;60    }61    sort(A, n);62    n = unique(A, n);63    int ans = 0;64    for (int i = 0; i < n; i++)65        for (int j = 0; j < n; j++)66            if (⑤ && binary_search(A, n, A[i].x, A[j].y) &&67                binary_search(A, n, A[j].x, A[i].y)) {68                ans++;69            }70    cout << ans << endl;71    return 0;72}