Problem:
Problem B: Electrophoretic

Scientist Frank, majoring in electrochemistry, has developed line-shaped strange electrodes called F-electrodes.
During being activated, each F-electrode causes a special potential on and between the two lines touching the
F-electrode’s endpoints at a right angle. Then electrically-charged particles located inside the potential area get
to move in the direction parallel to the potential boundary (i.e. perpendicular to the F-electrode), either toward or
against F-electrode. The moving direction can be easily controlled between the two possibles; it is also possible
to get particles to pass through F-electrodes. In addition, unlike ordinary electrodes, F-electrodes can affect
particles even infinitely far away, as long as those particles are located inside the potential area. On the other
hand, two different F-electrodes cannot be activated at a time, since their potentials conflict strongly.

We can move particles on our will by controlling F-electrodes. However, in some cases, we cannot lead them to
the desired positions due to the potential areas being limited. To evaluate usefulness of F-electrodes from some
aspect, Frank has asked you the following task: to write a program that finds the shortest distances from the
particles’ initial positions to their destinations with the given sets of F-electrodes.

Input

The input consists of multiple test cases. The first line of each case contains N (1 ≤ N ≤ 100) which represents
the number of F-electrodes. The second line contains four integers xs, ys, xt and yt, where (xs, ys) and (xt, yt)
indicate the particle’s initial position and destination. Then the description of N F-electrodes follow. Each line
contains four integers Fxs, Fys, Fxt and Fyt, where (Fxs, Fys) and (Fxt, Fyt ) indicate the two endpoints of an
F-electrode. All coordinate values range from 0 to 100 inclusive.

The input is terminated by a case with N = 0.

Output

Your program must output the case number followed by the shortest distance between the initial position to
the destination. Output “Impossible” (without quotes) as the distance if it is impossible to lead the elementary
particle to the destination. Your answers must be printed with five digits after the decimal point. No absolute
error in your answers may exceed 10-5.

Sample Input

2
2 1 2 2
0 0 1 0
0 1 0 2
0

Output for the Sample Input

Case 1: 3.00000

Solution in C (judge verdict: Accepted)

// AOJ 2113 Electrophoretic
// 2018.3.11 bal4u

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

#define QMAX 10000
typedef struct { int s; double t; } QUE;
QUE que[QMAX]; int qsize;

#define PARENT(i) ((i)>>1)
#define LEFT(i)   ((i)<<1)
#define RIGHT(i)  (((i)<<1)+1)

void min_heapify(int i)
{
	int l, r, min;
	QUE qt;

	l = LEFT(i), r = RIGHT(i);
	if (l < qsize && que[l].t < que[i].t) min = l; else min = i;
	if (r < qsize && que[r].t < que[min].t) min = r;
	if (min != i) {
		qt = que[i], que[i] = que[min], que[min] = qt;
		min_heapify(min);
	}
}

void deq()
{
	que[0] = que[--qsize];
	min_heapify(0);
}

void enq(int s, double t)
{
	int i, min;
	QUE qt;

	i = qsize++;
	que[i].s = s, que[i].t = t;
	while (i > 0 && que[min = PARENT(i)].t > que[i].t) {
		qt = que[i], que[i] = que[min], que[min] = qt;
		i = min;
	}
}

#define N			30000
#define INF			1e8
#define EPS			1e-5
#define EQ(a,b)		(fabs((a)-(b))<EPS)

typedef struct { double x, y; } PP;
typedef struct { PP s, e; } SEG, LINE;

SEG    trode[102];
LINE   line[410];
PP tbl [N]; int sz;
short  u[N];
short  hi[N], lim[N], *to[N];
char   mk[N];

#define PPeQ(a,b)	(EQ(a.x,b.x)&&EQ(a.y,b.y))
PP vset(double x, double y) { PP r; r.x = x, r.y = y; return r; }
PP vadd(PP p1, PP p2) { PP r; r.x = p1.x + p2.x, r.y = p1.y + p2.y; return r; }
PP vsub(PP p1, PP p2) { PP r; r.x = p1.x - p2.x, r.y = p1.y - p2.y; return r; }
PP vMul(PP p, double k) { PP r; r.x = p.x * k, r.y = p.y * k; return r; }
PP vsmul(PP p, double k) { PP r; r.x = p.x * k, r.y = p.y * k; return r; }
PP vmul(PP p1, PP p2) { PP r;
	r.x = p1.x * p2.x - p1.y * p2.y, r.y = p1.x * p2.y + p1.y * p2.x; return r; }

double cross(PP a, PP b) { return a.x * b.y - a.y * b.x; }
double dot(PP a, PP b) { return a.x * b.x + a.y * b.y; }
double norm(PP a) { return a.x * a.x + a.y * a.y; }
double vabs(PP a) { return hypot(a.x, a.y); }
int ccw(PP p0, PP p1, PP p2) { PP a, b;	double t;
	a = vsub(p1, p0), b = vsub(p2, p0), t = cross(a, b);
	if (t >  EPS) return 1;	if (t < -EPS) return -1;
	if (dot(a, b) < -EPS) return 2;	if (norm(a) < norm(b)) return -2;
	return 0;
}
PP dir(SEG s) { return vsub(s.e, s.s); }

int intersect(SEG s1, SEG s2)
{
	return ccw(s1.s, s1.e, s2.s) * ccw(s1.s, s1.e, s2.e) <= 0 &&
		   ccw(s2.s, s2.e, s1.s) * ccw(s2.s, s2.e, s1.e) <= 0;
}

PP crossPoint(int *ok, SEG s1, SEG s2)
{
	double n, d;
	PP one = {1,1};
	n = cross(vsub(s2.s, s1.s), dir(s2));
	d = cross(dir(s1), dir(s2));
	if (EQ(d,0)) { *ok = 0; return one; } 
	*ok = 1;
	return vadd(s1.s, vsmul(dir(s1), n/d));
}

#define BASE 100
// sに直交なLine2本
void orthogonalLL(LINE *ln1, LINE *ln2, SEG s)
{
	ln1->s = vadd(s.s, vmul(vsub(s.s, s.e), vset(0, BASE)));
	ln1->e = vadd(s.s, vmul(vsub(s.s, s.e), vset(0,-BASE)));
	ln2->s = vadd(s.e, vmul(vsub(s.s, s.e), vset(0, BASE)));
	ln2->e = vadd(s.e, vmul(vsub(s.s, s.e), vset(0,-BASE)));
}

void orthogonalL(LINE *ln, SEG s, PP p)
{
	ln->s = vadd(p, vmul(vsub(s.s, s.e), vset(0, BASE)));
	ln->e = vadd(p, vmul(vsub(s.s, s.e), vset(0,-BASE)));
}

//#define getchar_unlocked()  getchar()
int in()
{
	int n = 0;
	int c = getchar_unlocked();
	do n = 10*n + (c & 0xf), c = getchar_unlocked();
	while (c >= '0');
	return n;
}

int cmp(PP *a, PP *b) {
	if (EQ(a->x, b->x)) {
		if (EQ(a->y, b->y)) return 0;
		if (a->y <= b->y) return -1;
		return 1;
	}
	if (a->x <= b->x) return -1;
	return 1;
}

int uniq(PP *a, int n)
{
	int i, j;
	
	if (n == 0) return 0;
	for (i = 0, j = 1; j < n; j++) {
		while (j < n && PPeQ(a[i], a[j])) j++;
		if (j < n) if (++i != j) a[i] = a[j];
	}
	return i+1;
}

void check_lim(int i)
{
	if (hi[i] < lim[i]) return;
	if (lim[i] == 0) {
		lim[i] = 16, hi[i] = 0;			// 16 ?
		to[i] = malloc(sizeof(short) * lim[i]);
	} else {
		int k = lim[i];
		if (k > 128) k = 128;
		lim[i] += k;
		to[i] = realloc(to[i], sizeof(short) * lim[i]);
	}
}

void pushBack(int i, int j)
{
	int k;

	for (k = 0; k < hi[i]; k++) if (to[i][k] == j) return;
	check_lim(i);
	to[i][hi[i]++] = j;
}

void segmentArrangement(int n, PP start, PP goal)
{
	int i, j, k;
	
	tbl[0] = start, tbl[1] = goal, sz = 2;
	for (i = 0; i < n; i++) for (j = 0; j < i; j++) {
		if (intersect(line[i], line[j])) tbl[sz++] = crossPoint(&k, line[i], line[j]);
	}

	qsort(tbl, sz, sizeof(PP), cmp);
	sz = uniq(tbl, sz);

	memset(hi, 0, sz << 1), memset(lim, 0, sz << 1);
	for (i = 0; i < n; i++) {
		k = 0;
		for (j = 0; j < sz; j++) {
			if (!ccw(line[i].s, line[i].e, tbl[j])) u[k++] = j;
		}
		for (j = 0; j < k-1; j++) pushBack(u[j], u[j+1]), pushBack(u[j+1], u[j]);
	}
}

double dijkstra(PP start, PP goal)
{
	int i, s, e, sno, gno;
	double t, nt;

	sno = gno = -1;
	for (i = 0; i < sz; i++) {
		if (sno < 0 && PPeQ(tbl[i], start)) sno = i;
		if (gno < 0 && PPeQ(tbl[i], goal )) gno = i;
		if (sno >= 0 && gno >= 0) break;
	}

	memset(mk, 0, sz);
	qsize = 0;
	enq(sno, 0);
	while (qsize) {
		s = que[0].s, t = que[0].t, deq();
		if (s == gno) return t;

		if (mk[s]) continue;
		mk[s] = 1;

		for (i = 0; i < hi[s]; i++) {
			e = to[s][i], nt = t + vabs(vsub(tbl[s], tbl[e]));
			if (mk[e]) continue;
			enq(e, nt);
		}
	}
	return -1;
}

int main()
{
	int cno, n, i, m;
	double ans;
	PP s, t;

	cno = 0; while (n = in()) {

		s.x = in(), s.y = in();
		t.x = in(), t.y = in();

		tbl[0] = s, tbl[1] = t, sz = 2;

		for (i = 0; i < n; i++) {
			trode[i].s.x = in(), trode[i].s.y = in();
			trode[i].e.x = in(), trode[i].e.y = in();
		}

		m = 0;
		for (i = 0; i < n; i++) {
			orthogonalLL(line+m, line+m+1, trode[i]), m += 2;
			orthogonalL(line+m, trode[i], s);
			if (intersect(trode[i], line[m])) m++;
			orthogonalL(line+m, trode[i], t);
			if (intersect(trode[i], line[m])) m++;
		}

		segmentArrangement(m, s, t);

		ans = dijkstra(s, t);
		printf("Case %d: ", ++cno);
		if (ans < 0) puts("Impossible");
		else printf("%.5lf\n", ans);

		for (i = 0; i < sz; i++) if (lim[i]) free(to[i]);
	}
	return 0;
}