by Leap_Frog @ 2021-10-18 21:34:06
~~平面最近点对加强加强版的加强版~~
by Carnival @ 2021-10-18 21:35:25
```
return 1.0 / 0.0;
```
这不是无穷大吗？
by cancan123456 @ 2021-10-18 21:35:38
样例 1 输出了 200000000000000.
by cancan123456 @ 2021-10-18 21:36:32
看走眼了，dbq
by Leap_Frog @ 2021-10-18 21:37:02
orz
by bigju @ 2021-10-18 21:38:16
~~除数为0不会RE吗~~
by ScottSuperb @ 2021-10-18 21:58:26
我的情况跟你一样。。
[关于 EVOI Round 2 C题 的疑惑](https://www.luogu.com.cn/discuss/367230)
by 大K @ 2021-10-18 21:58:40
你说得对，样例加进数据了。
不过其实本来也只是挂了一个小边界？
by fstqwq @ 2021-10-24 18:10:46
我找到问题了，确实是边界挂了，这么写就对了：
```cpp
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
struct Point {
double x, y;
} point[400005];
bool operator < (const Point & a, const Point & b) {
return a.x == b.x ? a.y < b.y : a.x < b.x;
}
bool cmpy(const int & a, const int & b) {
return point[a].y < point[b].y;
}
double inline dist(double x1, double y1, double x2, double y2) {
return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}
double inline dist(int i, int j) {
return dist(point[i].x, point[i].y, point[j].x, point[j].y);
}
double inline min(double a, double b) {
return a < b ? a : b;
}
int temp[400005];
double calc(int l, int r) {
if (l == r - 1) {
return 1.0 / 0.0;
}
double d = 1.0 / 0.0;
int mid = (l + r) / 2;
d = min(d, min(calc(l, mid), calc(mid, r)));
int len = 0;
for (int i = l; i < r; i++) {
if (fabs(point[i].x - point[mid].x) < d) {
temp[len ++] = i;
}
}
sort(temp, temp + len, cmpy);
for (int i = 0; i < len; i++) {
for (int j = i + 1; j < len && point[temp[j]].y - point[temp[i]].y < d; j++) {
d = min(d, dist(temp[i], temp[j]));
}
}
return d;
}
int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%lf %lf", &point[i].x, &point[i].y);
}
sort(point, point + n);
double ans = calc(0, n);
printf("%.0lf", ans * ans);
return 0;
}
```
by cancan123456 @ 2021-11-15 20:45:56