题目链接：https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=0&problem=3892&mosmsg=Submission+received+with+ID+26561660

如果边按边权的排名不变，那么最小生成树的边就不会改变，发生改变的时刻，一定有两条边边长相等

涉及解二次方程组（很重要，细节注意）

所以找出所有边长相等的时间点，并且旧边是在当前最小生成树中，新边不在，则最小生成树发生改变，重新构建最小生成树

（以后自己独立完成一遍）

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn = 55;
const double eps = 1e-8;

int n, ans;

double x[maxn], y[maxn], z[maxn], vx[maxn], vy[maxn], vz[maxn];

int cnt = 0;
struct Edge{
	double a, b, c;
	int u, v;
	
	bool operator < (const Edge& x) const {
		return c - x.c < 0;
	}
}e[maxn * maxn];

struct Event{
	double t;
	int newe, olde;
	
	Event(double t=0, int newe=0, int olde=0) : t(t), newe(newe), olde(olde) {}
	
	bool operator < (const Event& x) const {
		return t - x.t < 0;
	}
};

vector<Event> event;

double sqr(double x) {
	return x * x;
}

void make_edges(){
	cnt = 0;
	for(int i = 1 ; i <= n ; ++i){
		for(int j = i + 1 ; j <= n ; ++j){
			double a = sqr(vx[i] - vx[j]) + sqr(vy[i] - vy[j]) + sqr(vz[i] - vz[j]);
			double b = 2 * (x[i] - x[j]) * (vx[i] - vx[j]) + 2 * (y[i] - y[j]) * (vy[i] - vy[j]) + 2 * (z[i] - z[j]) * (vz[i] - vz[j]);
			double c = sqr(x[i] - x[j]) + sqr(y[i] - y[j]) + sqr(z[i] - z[j]);
			++cnt;
			e[cnt].a = a; e[cnt].b = b; e[cnt].c = c; e[cnt].u = i; e[cnt].v = j;
		}
	}
	sort(e + 1, e + 1 + cnt);
}

void make_event(){
	event.clear();
	for(int i = 1 ; i <= cnt ; ++i){
		for(int j = i + 1 ; j <= cnt ; ++j){
			int s1 = i, s2 = j;
			if(e[s1].a - e[s2].a < 0) s1 = j, s2 = i;
			
			double a = e[s1].a - e[s2].a;
			double b = e[s1].b - e[s2].b;
			double c = e[s1].c - e[s2].c;
			
			if(fabs(a) < eps){ // bt + c = 0
				if(fabs(b) < eps) continue;
				if(b > 0){ swap(s1, s2); b = -b, c = -c; }
				if(c > 0) event.push_back(Event(-c / b, s1, s2));
				continue;
			}
			
			double delta = b * b - 4 * a * c;
			if(delta < eps) continue;
			delta = sqrt(delta);
			
			double t1 = -(b + delta) / (2 * a); // solution 1 
			double t2 = (delta - b) / (2 * a); // solution2 
			
			if(t1 > 0) event.push_back(Event(t1, s1, s2));
			if(t2 > 0) event.push_back(Event(t2, s2, s1));
		} 
	}
	sort(event.begin(), event.end());
}

int par[maxn];
void init_dsu(){
	for(int i = 1 ; i <= n ; ++i) par[i] = i;
}
int find(int x){
	return par[x] == x ? x : par[x] = find(par[x]);
}

int pos[maxn * maxn], re[maxn];
void solve(){
	init_dsu();
	memset(pos, 0, sizeof(pos));
	memset(re, 0, sizeof(re));
	int tot = 0;
	
	for(int i = 1 ; i <= cnt ; ++i){ // initial MST
		int u = find(e[i].u), v = find(e[i].v);
		if(u != v){
			par[u] = v;
			++tot;
			pos[i] = tot;
			re[tot] = i;
		}
		if(tot == n - 1) break;
	}
	
	ans = 1;
	for(int i = 0 ; i < event.size() ; ++i){
		if(pos[event[i].olde] && (!pos[event[i].newe])){
			int oldpos = pos[event[i].olde];
			
			// rebuild MST 
			init_dsu();
			
			for(int j = 1 ; j < n ; ++j){
				if(j != oldpos){
					int u = find(e[re[j]].u), v = find(e[re[j]].v);
					if(u != v){
						par[u] = v;
					}					
				}
			} 
			
			int u = find(e[event[i].newe].u), v = find(e[event[i].newe].v);
			if(u != v){ // find new MST
				par[u] = v;
				++ans;
				pos[event[i].newe] = oldpos;
				re[oldpos] = event[i].newe;
				pos[event[i].olde] = 0;
			}
		}
	}
}

ll read(){ ll s = 0, f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); } while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); } return s * f; }

int main(){
	int kase = 0;
	while(scanf("%d", &n) == 1){
		cnt = 0;
		
		for(int i = 1 ; i <= n ; ++i){
			scanf("%lf%lf%lf%lf%lf%lf", &x[i], &y[i], &z[i], &vx[i], &vy[i], &vz[i]);
		}
		
		make_edges();
		
		make_event();
		
		solve();
		
		printf("Case %d: %d\n", ++kase, ans);
	}
	return 0;
}