Tinkoff Challenge - Elimination Round C. Mice problem

C. Mice problem

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Igor the analyst fell asleep on the work and had a strange dream. In the dream his desk was crowded with computer mice, so he bought a mousetrap to catch them.

The desk can be considered as an infinite plane, then the mousetrap is a rectangle which sides are parallel to the axes, and which opposite sides are located in points (x1, y1) and (x2, y2).

Igor wants to catch all mice. Igor has analysed their behavior and discovered that each mouse is moving along a straight line with constant speed, the speed of the i-th mouse is equal to (vix, viy), that means that the x coordinate of the mouse increases by vix units per second, while the y coordinates increases by viy units. The mousetrap is open initially so that the mice are able to move freely on the desk. Igor can close the mousetrap at any moment catching all the mice that are strictly inside the mousetrap.

Igor works a lot, so he is busy in the dream as well, and he asks you to write a program that by given mousetrap's coordinates, the initial coordinates of the mice and their speeds determines the earliest time moment in which he is able to catch all the mice. Please note that Igor can close the mousetrap only once.

Input

The first line contains single integer n (1 ≤ n ≤ 100 000) — the number of computer mice on the desk.

The second line contains four integers x1, y1, x2 and y2 (0 ≤ x1 ≤ x2 ≤ 100 000), (0 ≤ y1 ≤ y2 ≤ 100 000) — the coordinates of the opposite corners of the mousetrap.

The next n lines contain the information about mice.

The i-th of these lines contains four integers rix, riy, vix and viy, (0 ≤ rix, riy ≤ 100 000,  - 100 000 ≤ vix, viy ≤ 100 000), where (rix, riy) is the initial position of the mouse, and (vix, viy) is its speed.

Output

In the only line print minimum possible non-negative number t such that if Igor closes the mousetrap at t seconds from the beginning, then all the mice are strictly inside the mousetrap. If there is no such t, print -1.

Your answer is considered correct if its absolute or relative error doesn't exceed 10 - 6.

Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if .

Examples

input

4
7 7 9 8
3 5 7 5
7 5 2 4
3 3 7 8
6 6 3 2

output

0.57142857142857139685

input

4
7 7 9 8
0 3 -5 4
5 0 5 4
9 9 -1 -6
10 5 -7 -10

output

-1

Note

Here is a picture of the first sample

Points A, B, C, D - start mice positions, segments are their paths.

Then, at first time when all mice will be in rectangle it will be looks like this:

Here is a picture of the second sample

Points A, D, B will never enter rectangle.

 

题目大意：给出坐标系中的一个矩形和一些点和他们的速度向量，求一个最小的时刻t使得这些点都严格在矩形内

 

这个严格要害死人哦

看上去只要枚举四条边求一下相交时间取个交集就好，然而特殊情况特别多

首先考虑一些和矩形相交在角上的坑爹情况，然后还有什么速度是0，在边上运动，最后还有答案区间是个(0.5,0.5)这种情况

一个很好的办法是求出相交时间后向正和向负各松弛eps，如果有一边严格在矩形内部就认为这个解合法

最后还要再判一下r-l是不是大于eps

wa了12次

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#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cstring>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <vector>
#include <queue>
#include <time.h>
#define eps 1e-10
#define INF 0x3f3f3f3f
#define MOD 1000000007
#define rep0(j,n) for(int j=0;j<n;++j)
#define rep1(j,n) for(int j=1;j<=n;++j)
#define pb push_back
#define set0(n) memset(n,0,sizeof(n))
#define ll long long
#define ull unsigned long long
#define iter(i,v) for(edge *i=head[v];i;i=i->nxt)
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
#define print_rumtime printf("Running time:%.3lfs\n",(long double)(clock())/1000.0);
#define TO(j) printf(#j": %d\n",j);
//#define OJ
using namespace std;
const int MAXINT = 100010;
const int MAXNODE = 100010;
const int MAXEDGE = 2 * MAXNODE;
char BUF, *buf;
int read() {
    char c = getchar(); int f = 1, x = 0;
    while (!isdigit(c)) { if (c == '-') f = -1; c = getchar(); }
    while (isdigit(c)) { x = x * 10 + c - '0'; c = getchar(); }
    return f * x;
}
char get_ch() {
    char c = getchar();
    while (!isalpha(c)) c = getchar();
    return c;
}
//------------------- Head Files ----------------------//
int x_s, x_l, y_s, y_l, n;
long double lb = -1e50, rb = 1e50;
struct mouse {
    int vx, vy, sx, sy;
    mouse(int _sx, int _sy, int _vx, int _vy) {
        sx = _sx; sy = _sy; vx = _vx; vy = _vy;
    }
    mouse() {}
    long double intersect(int p, int tp) {
        long double t;
        if (tp == 1) { //x=p;
            t = ((long double)(p)-sx) / vx;
            long double py1 = sy + vy * (t + eps);
            long double py2 = sy + vy * (t - eps);
            if (((t + eps > 0) && (py1<y_l && py1>y_s)) || ((t - eps > 0) && (py2<y_l && py2>y_s))) return t;
            return -1;
        }
        if (tp == 2) { //y=p
            t = ((long double)(p)-sy) / vy;
            long double px1 = sx + vx * (t + eps);
            long double px2 = sx + vx * (t - eps);
            if (((t + eps > 0) && (px1<x_l && px1>x_s)) || ((t - eps > 0) && (px2<x_l && px2>x_s))) return t;
            return -1;
        }
    }
    int in() {
        return (sx < x_l && sx > x_s) && (sy < y_l && sy > y_s);
    }
} mice[MAXINT];
int intersect(long double l, long double r) {
    lb = max(l, lb);
    rb = min(r, rb);
    if (rb - lb < eps) return 0;
    return 1;
}
void get_input();
void work();
int main() {
    get_input();
    work();
    return 0;
}
void work() {
    rep0(i, n) {
        long double t, l = 1e50, r = -1e50;
        if (mice[i].in()) {
            l = 0;
            if (mice[i].vx == 0 && mice[i].vy == 0) r = 1e50;
        }
        if ((t = mice[i].intersect(x_s, 1)) > -eps) l = min(l, t), r = max(r, t);
        if ((t = mice[i].intersect(x_l, 1)) > -eps) l = min(l, t), r = max(r, t);
        if ((t = mice[i].intersect(y_s, 2)) > -eps) l = min(l, t), r = max(r, t);
        if ((t = mice[i].intersect(y_l, 2)) > -eps) l = min(l, t), r = max(r, t);
        if (r - l < eps) {
            printf("-1\n");
            return;
        }
        if (!intersect(l, r)) {
            printf("-1\n");
            return;
        }
    }
    printf("%.10Lf\n", lb);
}
void get_input() {
    n = read();
    int x1 = read(), y1 = read(), x2 = read(), y2 = read();
    x_s = min(x1, x2);
    x_l = max(x1, x2);
    y_s = min(y1, y2);
    y_l = max(y1, y2);
    rep0(i, n) {
        mice[i].sx = read();
        mice[i].sy = read();
        mice[i].vx = read();
        mice[i].vy = read();
    }
}