(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 <iostream>





using namespace std;





struct point {


	int x, y, id;


};





bool equals(point a, point b) {


	return a.x == b.x && a.y == b.y;


}





bool cmp(point a, point b) {


	return ___(1)___;


}





void sort(point A[], int n) {


	for(int i=0; i<n; i++)


		for(int j=1; j<n; j++)


			if (cmp(A[j], A[j - 1])) {


				point t = A[j];


				A[j] = A[j - 1];


				A[j-1]=t;


			}


}





int unique(point A[], int n) {


	int t=0;


	for(int i=0; i<n; i++)


		if (___(2)___)


			A[t++] = A[i];


	return t;


}





bool binary_search(point A[], int n, int x, int y) {


	point p;


	p.x = x;


	p.y =y;


	p.id = n;


	int a=0,b=n-1;


	while(a<b) {


		int mid = ___(3)___;


		if (___(4)___)


			a=mid+1;


		else


			b = mid;


	}


	return equals(A[a], p);


}





const int MAXN = 1000;


point A[MAXN];





int main() {


	int n;


	cin >> n;


	for(int i=0; i<n; i++) {


		cin >> A[i].x >> A[i].y;


		A[i].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 ( ___(5)___ && binary_search(A, n, A[i].x, A[j].y) && binary_search(A, n, A[j].x, A[i].y)) {


				ans++;


			}


	cout<<ans<<endl;


	return 0;


}

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