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(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 ①;
}
void sort(struct 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])) {
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 (②)
A[t++] = A[i];
return t;
}
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 = ③;
if (④)
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[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++;
}
printf("%d\n", ans);
return 0;
}
①处应填()
②处应填()
③处应填()
④处应填()
⑤处应填()
1.
A.a.x != b.x ? a.x < b.x : a.id < b.id
B.a.x != b.x ? a.x < b.x : a.y < b.y
C.equals(a,b) ? a.id < b.id : a.x < b.x
D.equals(a,b) ? a.id < b.id : (a.x != b.x ? a.x < b.x : a.y < b.y)
2.
A.i == 0 || cmp(A[i], A[i - 1])
B.t == 0 || equals(A[i], A[t - 1])
C.i == 0 || !cmp(A[i], A[i - 1])
D.t == 0 || !equals(A[i], A[t - 1])
3.
A.b - (b - a) / 2 + 1
B.(a + b + 1) >> 1
C.(a + b) >> 1
D.a + (b - a + 1) / 2
4.
A.!cmp(A[mid], p)
B.cmp(A[mid], p)
C.cmp(p, A[mid])
D.!cmp(p, A[mid])
5.
A.A[i].x == A[j].x
B.A[i].id < A[j].id
C.A[i].x == A[j].x && A[i].id < A[j].id
D.A[i].x < A[j].x && A[i].y < A[j].y
第一空(3分):_______________________
第二空(3分):_______________________
第三空(3分):_______________________
第四空(3分):_______________________
第五空(3分):_______________________