首页 > ACM题库 > HDU-杭电 > HDU 1445 Ride to School-贪心-[解题报告] C++
2013
12-10

HDU 1445 Ride to School-贪心-[解题报告] C++

Ride to School

问题描述 :

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.

输入:

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.

输出:

Output one line for each case: the arrival time of Charley. Round up (ceiling) the value when dealing with a fraction.

样例输入:

4
20 0
25 -155
27 190
30 240
2
21 0
22 34
0

样例输出:

780
771

http://acm.hdu.edu.cn/showproblem.php?pid=1445

分析:原先把问题想麻烦了,其实只要理清思路就会发现,无论自己的车速会怎么变,最后一定会和一个人同时到达,那么计算这个人的到达时间即可
其中,时间为负数的不用考虑,要么你追不上,要么他追不上你

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;

const int NM=10005;
const int SEC=3600;
const double D=4.5;
struct Ride{
	int vi,ti;
}r[NM];

bool comp(struct Ride A,struct Ride B)
{
	if(A.ti<B.ti) return 1;
	else
	{
		if(A.ti==B.ti)
		{
			if(A.vi<B.vi) return 0;
			else return 0;
		}
		else return 0;
	}
}

int main()
{
	int n,i,j;
	double time1,time;
	while(scanf("%d",&n)&&n)
	{
		for(i=0;i<n;i++)
			scanf("%d%d",&r[i].vi,&r[i].ti);
		sort(r,r+n,comp);
		for(j=0;j<n;j++)  //<0 忽略
		{
			if(r[j].ti>=0)
			{
				time=(double)(D*SEC/r[j].vi)+r[j].ti;
				break;
			}
		}
	
		for(i=j+1;i<n;i++)
		{
			time1=(double)(D*SEC/r[i].vi)+r[i].ti;
			if(time1<time)
				time=time1;
		}
		
		printf("%.lf\n",ceil(time));  //返回不小于这个数的最小整数
	}
	return 0;
}

解题报告转自:http://blog.csdn.net/killua_99/article/details/9457249


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