C. Mice problem
time limit per test
2 secondsmemory limit per test
256 megabytesinput
standard inputoutput
standard outputIgor 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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LBQrsZRBMBQBQRBBQBBRGXPP. .,,,,,,,.:7. ,7GOEOgDQZJ66HPG1LUPJJ5asaEJ:.,,,,,,,,,,,,. 91 //: :,,, 5QJ ,,::r7sLsvL7r;:. .; ..,,,,,,,,.:r .. :cZRBBBBDaK6GLJXHLJX1JDXs..,,,,,,,,,,,:. 92 //,.,:,. 1BO,..,,. .r..,,,:,,,,,:.:r.,: ,J2OEHXDassEUJUKLXDXr .,,,,,,,,,,:,. 93 //:.:,, wBB;..::: :7,,,:,:,:,:,:,,:L.::, sBSPO1swG5sKULpO2:.:,:,:,:,:,:,:. 94 //: .. :Bs .., ,; ............. .r .., ,.wBGK5rL2EJs5scP57 ............. 95 96 97 #include <iostream> 98 #include <cstdlib> 99 #include <cstdio> 100 #include <algorithm> 101 #include <string> 102 #include <cstring> 103 #include <cmath> 104 #include <map> 105 #include <stack> 106 #include <set> 107 #include <vector> 108 #include <queue> 109 #include <time.h> 110 #define eps 1e-10 111 #define INF 0x3f3f3f3f 112 #define MOD 1000000007 113 #define rep0(j,n) for(int j=0;j<n;++j) 114 #define rep1(j,n) for(int j=1;j<=n;++j) 115 #define pb push_back 116 #define set0(n) memset(n,0,sizeof(n)) 117 #define ll long long 118 #define ull unsigned long long 119 #define iter(i,v) for(edge *i=head[v];i;i=i->nxt) 120 #define max(a,b) (a>b?a:b) 121 #define min(a,b) (a<b?a:b) 122 #define print_rumtime printf("Running time:%.3lfs ",(long double)(clock())/1000.0); 123 #define TO(j) printf(#j": %d ",j); 124 //#define OJ 125 using namespace std; 126 const int MAXINT = 100010; 127 const int MAXNODE = 100010; 128 const int MAXEDGE = 2 * MAXNODE; 129 char BUF, *buf; 130 int read() { 131 char c = getchar(); int f = 1, x = 0; 132 while (!isdigit(c)) { if (c == '-') f = -1; c = getchar(); } 133 while (isdigit(c)) { x = x * 10 + c - '0'; c = getchar(); } 134 return f * x; 135 } 136 char get_ch() { 137 char c = getchar(); 138 while (!isalpha(c)) c = getchar(); 139 return c; 140 } 141 //------------------- Head Files ----------------------// 142 int x_s, x_l, y_s, y_l, n; 143 long double lb = -1e50, rb = 1e50; 144 struct mouse { 145 int vx, vy, sx, sy; 146 mouse(int _sx, int _sy, int _vx, int _vy) { 147 sx = _sx; sy = _sy; vx = _vx; vy = _vy; 148 } 149 mouse() {} 150 long double intersect(int p, int tp) { 151 long double t; 152 if (tp == 1) { //x=p; 153 t = ((long double)(p)-sx) / vx; 154 long double py1 = sy + vy * (t + eps); 155 long double py2 = sy + vy * (t - eps); 156 if (((t + eps > 0) && (py1<y_l && py1>y_s)) || ((t - eps > 0) && (py2<y_l && py2>y_s))) return t; 157 return -1; 158 } 159 if (tp == 2) { //y=p 160 t = ((long double)(p)-sy) / vy; 161 long double px1 = sx + vx * (t + eps); 162 long double px2 = sx + vx * (t - eps); 163 if (((t + eps > 0) && (px1<x_l && px1>x_s)) || ((t - eps > 0) && (px2<x_l && px2>x_s))) return t; 164 return -1; 165 } 166 } 167 int in() { 168 return (sx < x_l && sx > x_s) && (sy < y_l && sy > y_s); 169 } 170 } mice[MAXINT]; 171 int intersect(long double l, long double r) { 172 lb = max(l, lb); 173 rb = min(r, rb); 174 if (rb - lb < eps) return 0; 175 return 1; 176 } 177 void get_input(); 178 void work(); 179 int main() { 180 get_input(); 181 work(); 182 return 0; 183 } 184 void work() { 185 rep0(i, n) { 186 long double t, l = 1e50, r = -1e50; 187 if (mice[i].in()) { 188 l = 0; 189 if (mice[i].vx == 0 && mice[i].vy == 0) r = 1e50; 190 } 191 if ((t = mice[i].intersect(x_s, 1)) > -eps) l = min(l, t), r = max(r, t); 192 if ((t = mice[i].intersect(x_l, 1)) > -eps) l = min(l, t), r = max(r, t); 193 if ((t = mice[i].intersect(y_s, 2)) > -eps) l = min(l, t), r = max(r, t); 194 if ((t = mice[i].intersect(y_l, 2)) > -eps) l = min(l, t), r = max(r, t); 195 if (r - l < eps) { 196 printf("-1 "); 197 return; 198 } 199 if (!intersect(l, r)) { 200 printf("-1 "); 201 return; 202 } 203 } 204 printf("%.10Lf ", lb); 205 } 206 void get_input() { 207 n = read(); 208 int x1 = read(), y1 = read(), x2 = read(), y2 = read(); 209 x_s = min(x1, x2); 210 x_l = max(x1, x2); 211 y_s = min(y1, y2); 212 y_l = max(y1, y2); 213 rep0(i, n) { 214 mice[i].sx = read(); 215 mice[i].sy = read(); 216 mice[i].vx = read(); 217 mice[i].vy = read(); 218 } 219 }