2020 ICPC Asia Taiwan Online Programming Contest C Circles

This problem really ignores the process and will be wrong , In the future, I will not write directly with any ideas .
Topic link ++++++++++++++++++++++++
&&&&& I have just given you n Center of circle , After the game starts, the radius of each circle begins to increase at the same rate , When one circle meets another circle, the circle stops growing . In this case , As the circle grows, the radius of other circles will be limited （ This is dynamic , The smallest two circles must stop growing first , Then the other center radii change , Find the minimum value each time , You can use priority queues ）, It is not to find the center nearest to each center that is where the growth stops ,
The title is as follows
There are n magical circles on a plane. They are centered at (x1, y1),(x2, y2), . . . ,(xn, yn), respectively. In the beginning, the radius of each circle is 0, and the radii of all magical circles will grow at the same rate. When an magical circle touches another, then it stops growing. Write a program to calculate the total area of all magical circles at the end of growing.

Input description :
The first line contains an integer n to indicate the number of magical circles. The i-th of thefollowing n lines contains two space-separated integers xi and yi indicating that the i-th magicalcircle is centered at (xi , yi).
Output description :
Output the total area of the circles.

Input
Copy
3
0 0
0 1
2 0
Output
Copy
8.639379797371932

Input
Copy
4
0 0
1 0
1 1
0 1
Output
Copy
3.14159265359

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 2010;
const ll MOD = 1000000007;
const double EPS = 1e-8;
const double pi = acos(-1.0);
 
struct Node
{
    
    double dis;
    int a, b;
    bool operator < (const Node &x) const
    {
    
        return dis > x.dis;
    }
};
priority_queue<Node> Q;
 
int v[MAXN];
double px[MAXN], py[MAXN], ans[MAXN];
 
inline double getdis(int a, int b)
{
    
    double X = abs(px[a] - px[b]);
    double Y = abs(py[a] - py[b]);
    return sqrt(X * X + Y * Y);
}
 
int main()
{
    
    int n; scanf("%d", &n);
    for(int i = 1; i <= n; ++i)
        scanf("%lf %lf", &px[i], &py[i]);
    for(int i = 1; i <= n; ++i)
        for(int j = i + 1; j <= n; ++j)
            Q.push((Node){
    getdis(i, j) / 2.0, i, j});
    while(!Q.empty())
    {
    
        Node x = Q.top(); Q.pop();
        if(v[x.a] && v[x.b]) continue;
        else if(!v[x.a] && !v[x.b])
        {
    
            v[x.a] = v[x.b] = 1;
            ans[x.a] = ans[x.b] = x.dis;
            for(int i = 1; i <= n; ++i)
            {
    
                if(!v[i]) Q.push((Node){
    getdis(x.a, i) - x.dis, x.a, i});
                if(!v[i]) Q.push((Node){
    getdis(x.b, i) - x.dis, x.b, i});
            }
        }
        else
        {
    
            if(v[x.a]) swap(x.a, x.b);
            if(abs(x.dis + ans[x.b] - getdis(x.a, x.b)) < EPS)
            {
    
                v[x.a] = 1, ans[x.a] = x.dis;
                for(int i = 1; i <= n; ++i)
                    if(!v[i]) Q.push((Node){
    getdis(x.a, i) - x.dis, x.a, i});
            }
        }
    }
    double sum = 0;
    for(int i = 1; i <= n; ++i)
        sum += ans[i] * ans[i] * pi;
    printf("%.15f\n", sum);
    return 0;
}

This question gives me a new way of thinking , frigging awesome ！！！！！！！！！