[SDOI2012]拯救小云公主

Posted 2022-12-22 kma093

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;