C - Safe Distance(二分&DSU)

显然答案具有二分性，然后就是套路性对每个点化圆。

考虑无解的情况，假设分成上下左右四个区域，显然上下连通或者左右连通无解，通路左下或右上连通无解。因此考虑对于相交的两圆进行并查集合并。

然后就裸了。

参考别人代码

#include 

using namespace std;

const int N = 1e3 + 10;

struct Res {
  double x, y;

  void read() {
    scanf("%lf %lf", &x, &y);
  }
}a[N];

int fa[N], n, X, Y; // {n + 1, 上表面} {n + 2, 下表面} {n + 3, 左表面} {n + 4, 右表面}

int find(int rt) {
  return rt == fa[rt] ? rt : fa[rt] = find(fa[rt]);
}

double Dis(Res a, Res b) {
  return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}

bool judge(double x) {
  for (int i = 1; i <= n + 4; i++) {
    fa[i] = i;
  }
  for (int i = 1; i <= n; i++) {
    if (a[i].y + x > Y) {
      int u = find(i), v = find(n + 1);
      fa[u] = v;
    }
    if (a[i].y < x) {
      int u = find(i), v = find(n + 2);
      fa[u] = v;
    }
    if (a[i].x < x) {
      int u = find(i), v = find(n + 3);
      fa[u] = v;
    }
    if (a[i].x + x > X) {
      int u = find(i), v = find(n + 4);
      fa[u] = v;
    }
    for (int j = 1; j < i; j++) {
      if (Dis(a[i], a[j]) < 2 * x) {
        int u = find(i), v = find(j);
        fa[u] = v;
      }
    }
  }
  if ((find(n + 1) == find(n + 2)) || (find(n + 1) == find(n + 4)) || (find(n + 3) == find(n + 2)) || (find(n + 3) == find(n + 4))) {
    return false;
  }
  return true;
} 

int main() {
  // freopen("in.txt", "r", stdin);
  // freopen("out.txt", "w", stdout);
  // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  scanf("%d %d %d", &X, &Y, &n);
  for (int i = 1; i <= n; i++) {
    a[i].read();
  }
  double l = 0, r = max(X, Y);
  for (int i = 1; i <= 100; i++) {
    double mid = 0.5 * (l + r);
    if (judge(mid)) {
      l = mid;
    }
    else {
      r = mid;
    }
  }
  printf("%.10fn", l);
  return 0;
}