#363. CSP2021PJ-20
CSP2021PJ-20
- (完善程序)
(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 _________________________;
}
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 (_________________________)
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 = _________________________;
if (_________________________)
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 (_________________________ && 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;
}
① 处应填( )。
{{ select(1) }}
- a.x != b.x ? a.x < b.x : a.id < b.id
- a.x != b.x ? a.x < b.x : a.y < b.y
- equals(a, b) ? a.id < b.id : a.x < b.x
- equals(a, b) ? a.id < b.id : (a.x != b.x ? a.x < b.x : a.y < b.y)
② 处应填( )。
{{ select(2) }}
- i == 0 || cmp(A[i], A[i - 1])
- t == 0 || equals(A[i], A[t - 1])
- i == 0 || !cmp(A[i], A[i - 1])
- t == 0 || !equals(A[i], A[t - 1])
③ 处应填( )。
{{ select(3) }}
- b - (b - a) / 2 + 1
- (a + b + 1) >> 1
- (a + b) >> 1
- a + (b - a + 1) / 2
④ 处应填( )。
{{ select(4) }}
- !cmp(A[mid], p)
- cmp(A[mid], p)
- cmp(p, A[mid])
- !cmp(p, A[mid])
⑤ 处应填( )。
{{ select(5) }}
- A[i].x == A[j].x
- A[i].id < A[j].id
- A[i].x == A[j].x && A[i].id < A[j].id
- A[i].x < A[j].x && A[i].y < A[j].y
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