POJ 1922 Ride to School

Ride to School Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 19069   Accepted: 7717 Description Many graduate students of Peking University are living in Wanliu Campus, which is 4.5 kilometers from the main campus – Yanyuan. Students in Wanliu have to either take a bus or ride a bike to go to school. Due to the bad traffic in Beijing, many students choose to ride a bike.  We may assume that all the students except "Charley" ride from Wanliu to Yanyuan at a fixed speed. Charley is a student with a different riding habit – he always tries to follow another rider to avoid riding alone. When Charley gets to the gate of Wanliu, he will look for someone who is setting off to Yanyuan. If he finds someone, he will follow that rider, or if not, he will wait for someone to follow. On the way from Wanliu to Yanyuan, at any time if a faster student surpassed Charley, he will leave the rider he is following and speed up to follow the faster one.  We assume the time that Charley gets to the gate of Wanliu is zero. Given the set off time and speed of the other students, your task is to give the time when Charley arrives at Yanyuan.  Input There are several test cases. The first line of each case is N (1 <= N <= 10000) representing the number of riders (excluding Charley). N = 0 ends the input. The following N lines are information of N different riders, in such format:  Vi [TAB] Ti  Vi is a positive integer <= 40, indicating the speed of the i-th rider (kph, kilometers per hour). Ti is the set off time of the i-th rider, which is an integer and counted in seconds. In any case it is assured that there always exists a nonnegative Ti.  Output Output one line for each case: the arrival time of Charley. Round up (ceiling) the value when dealing with a fraction. Sample Input 4 20 0 25 -155 27 190 30 240 2 21 0 22 34 0 Sample Output 780 771 Source Beijing 2004 Preliminary@POJ

题目大意：

同学们骑自行车去主校区上课，Charley有一个不同的骑车习惯——他总是要跟在一个骑车的同学后面，Charley在起点等待区主校区的同学，如果有骑车去主校区的同学，他就跟着这位同学，如果没有，他就继续等。在路上，如果有骑得更快的同学超过了Charley，他就离开原来的同学，跟着更快的同学走。

求Charley到达主校区的时间。Charley到达起点的时间是0.

解法：

求出每个同学到达主校区的时间，其中最少的就是Charley用的时间。（到达起点的时间大于0的不考虑，因为已经在Charley到达时走掉）

#include <iostream>
#include <math.h>
using namespace std;
int main()
{
    int n;
    while(cin>>n&&n)
    {
		double min=100000.0;
		int i;
		for(i=1;i<=n;i++)
		{
			double m;
			double v,t;
			cin>>v>>t;	
			if(t>=0)
			{
				m=t+4.5*3600/v;
				if(m<min)
					min=ceil(m);
			}
		}
		cout<<min<<endl;
    }
    return 0;
}