①处应填（）

（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。

#include <stdio.h>
struct point {
    int x, y, id;
};

int equals(struct point a, struct point b) {
    return a.x == b.x && a.y == b.y;
}

int cmp(struct point a, struct point b) {
    return 第一处;
}

void sort(struct point A[], int n) {
    for (int i = 0; i < n; i++)
        for (int j = i; j < n; j++)
            if (cmp(A[j], A[j - 1])) {
                struct point t = A[j];
                A[j] = A[j - 1];
                A[j - 1] = t;
            }
}

int unique(struct point A[], int n) {
    int t = 0;
    for (int i = 0; i < n; i++) {
        if (第二处))
            A[++t] = A[i];
    }
    return t + 1;
}

int binary_search(struct point A[], int n, int x, int y) {
    struct point p;
    p.x = x;
    p.y = y;
    p.id = n;
    int a = 0, b = n - 1;

    while(a < b) {
        int mid = 第三处;
        if (第四处)
            a = mid + 1;
        else
            b = mid;
    }
    return equals(A[a], p);
}

#define MAXN 1000
struct point A[MAXN];

int main() {
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
        scanf("%d %d", &A[i].x, &A[i].y);
    A[n].id = i;
    sort(A, n);
    n = unique(A, n);
    int ans = 0;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            if (i < j && binary_search(A, n, A[i].x, A[j].y) &&
                binary_search(A, n, A[j].x, A[i].y))
                ans++;
    printf("%d\n", ans);
    return 0;
}