UVALive - 2D-Solar System

题目链接：https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4760
Time limit: 3000 ms

Problem Description

The 2D-solar system like our solar system comprises Bigsun (its sun) and its planetary system of many circular planets orbiting around Bigsun. Due to the high gravity of Bigsun, all planets have been attracted by Bigsun. Precisely, they orbit around Bigsun while being tangent to it as depicted in the figure (As Bigsun is so huge, its boundary looks like a line.) Surprisingly, up to the current time no two planets have collided with each other, but no one knows whether the system is free of collisions
in the future. You are to write a program to verify whether there is a possibility of any collision in the future and if so, compute the time at which the first collision happens. The scientists of NASA have realized that each planet in the 2D-solar system moves with a constant velocity. More precisely, it turned out that the motion equation of a planet can be described by the position of its touching point with the boundary of Bigsun through time by the linear equation y=at+by = at + by=at+b where a and b are two known parameters and t denotes time.

Input

There are multiple test cases in the input. Each test case starts with a line containing an integer n(n≤n≤50,000)n (n ≤ n ≤ 50, 000)n(n≤n≤50,000) where nnn is the number of planets. The iii-th line of the next nnn lines contains 333 spaceseparated integers rir_iri, aia_iai, and bib_ibi whose absolute values are not exceeding 1,000,000,0001,000,000,0001,000,000,000. The number ririri which is a positive square number, denotes the radius of Planet iii and aia_iai and bib_ibi specify its motion equation, i.e. the position of the tangent point of the planet on the boundary of Bigsun at time ttt is ait+bia_it + b_iait+bi. The input terminates with a line containing ‘0’‘0’‘0’ which should not be processed.

Output

For each test case, output a line containing the time at which the first collision happens under the assumption that the current time is equal to 000 and all planets are disjoint at the current time. If the system is free of collisions you must output “Collision-Free System”. The output must be rounded to exactly two digits after the decimal point.

Sample Input

3
1 1 1
4 3 6
9 -7 30
2
4 -1 1
1 1 7
2
1 1 10
1 2 5
0

Sample Output

1.20
Collision-Free System
3.00

Problem solving report:

Description：有nnn个圆在一条直线上，直线无限长，每个圆有一个速度，可能往左，也可能往右，保证所有的圆在一开始的时候都相离。问能不能所有的圆永远都不相撞，如果相撞，输出相撞的时间。给出每个圆的半径rrr，速度aaa和起始位置bbb。
Problem solving：按照初始位置排序，那么答案必定小于等于相邻两个圆相切的情况。
注意到对于一个圆，令左端离他最近的半径比他大的圆的位置为LLL，右端离他最近的半径比他大的圆的位置为RRR，这个圆只会受到LLL到RRR这段区间的影响。
所以维护一个半径单调递减的栈即可。

Accepted Code:

/* * @Author: lzyws739307453 * @Language: C++ */
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 5e4 + 5;
const int inf = 0x3f3f3f3f;
struct Circle {int r, a, b;bool operator < (const Circle &s) const {return b < s.b;}
}p[MAXN];
int main() {int n;while (~scanf("%d", &n), n) {for (int i = 0; i < n; i++)scanf("%d%d%d", &p[i].r, &p[i].a, &p[i].b);sort(p, p + n);stack <Circle> S;S.push(p[0]);double min_ = inf;for (int i = 1; i < n; i++) {while (!S.empty()) {Circle q = S.top();if (q.a > p[i].a)min_ = min(min_, (p[i].b - q.b - 2 * sqrt(q.r * p[i].r)) / (q.a - p[i].a));if (q.r < p[i].r)S.pop();else break;}S.push(p[i]);}if (min_ < inf)printf("%.2lf\n", min_);else printf("Collision-Free System\n");}return 0;
}