[SDOI2012]拯救小云公主

题目链接：

题目

分析

最大的最小最小的最大 一看就非常的二分
考虑二分之后转化成判定联通问题
于是可以考虑奶酪
直接维护联通好像很(GG)，正难则反考虑维护不连通
什么时候不连通，发现当左上角和右下角分在两个块里面就凉了
当上和右边或上和下边联通的时候左上角和右下角被分割在两块
另外两种情况同理
于是并查集仿照奶酪维护一下，如果四种都没联通说明可以走到
外面再套个二分 复杂度(O(n ^ 2 * log(n))) 此代码吸氧可过

#include<bits/stdc++.h>
#define eps 1e-8
#define N (3000 + 10)
using namespace std;
inline int read() {
	int cnt = 0, f = 1; char c = getchar();
	while (!isdigit(c)) {if (c == '-') f = -f; c = getchar();}
	while (isdigit(c)) {cnt = (cnt << 3) + (cnt << 1) + (c ^ 48); c = getchar();}
	return cnt * f;
}
int n, R, L, fa[N];
double ans;
struct node {
	double x, y;
}a[N];
int get_fa(int x) {return x == fa[x] ? x : fa[x] = get_fa(fa[fa[fa[fa[x]]]]);}
double Pow(double x) {return x * x;}
bool dis(node a, node b, double r) {
	return (r * r * 4 - Pow(a.x - b.x) - Pow(a.y - b.y)) > 0;
}
bool check(double r) {
	for (register int i = 1; i <= n + 2; ++i) fa[i] = i;
	for (register int i = 1; i <= n; ++i) {
		if (a[i].x - r <= 1 || a[i].y + r >= L) fa[i] = n + 1;
		if (a[i].x + r >= R || a[i].y - r <= 1) fa[get_fa(i)] = n + 2;
	}
	for (register int i = 1; i <= n; ++i)
		for (register int j = i; j <= n; ++j) {
			int fx = get_fa(i), fy = get_fa(j);
			if (fx == fy) continue;
			if (!dis(a[i], a[j], r)) continue;
			fa[fx] = fy;
			if (get_fa(n + 1) == get_fa(n + 2)) return false;
		}
	if (get_fa(n + 1) == get_fa(n + 2)) return false;
	return true;
}
inline double binary() {
	double l = 0, r = min(L, R);
	double mid = (l + r) / 2.0;
	while (r - l > eps) {
//		cout<<l<<" "<<r<<endl;
		if (check(mid)) l = mid;
		else r = mid;
		mid = (l + r) / 2.0;
	}
	return mid;
}
int main() {
// 	freopen("1.in", "r", stdin);
// 	freopen("own.out", "w", stdout);
	n = read(), R = read(), L = read();
	for (register int i = 1; i <= n; ++i) a[i].x = read(), a[i].y = read();
	ans = binary();
	printf("%.2lf", ans);
	return 0;
}