C - Tram
Problem description
Linear Kingdom has exactly one tram line. It has n stops, numbered from 1 to n in the order of tram's movement. At the i-th stop ai passengers exit the tram, while bipassengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram.
Input
The first line contains a single number n (2 ≤ n ≤ 1000) — the number of the tram's stops.
Then n lines follow, each contains two integers ai and bi (0 ≤ ai, bi ≤ 1000) — the number of passengers that exits the tram at the i-th stop, and the number of passengers that enter the tram at the i-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally,
. This particularly means that a1 = 0.
- At the last stop, all the passengers exit the tram and it becomes empty. More formally,
.
- No passenger will enter the train at the last stop. That is, bn = 0.
Output
Print a single integer denoting the minimum possible capacity of the tram (0 is allowed).
Examples
Input
4
0 3
2 5
4 2
4 0
Output
6
Note
For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3.
- At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now.
- At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now.
- Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer.
解题思路:题目的意思就是找出上下车过程中车内的最多人数,即为车的至少容量,简单水过。题目已经保证了一开始下车的人数为0,最后上车的人数也为0,并且最后车内的人将全部下车。
AC代码:
#include<bits/stdc++.h>
using namespace std;
int main(){
int n,a,b,nowcap=,maxcap=;
cin>>n;
while(n--){
cin>>a>>b;
nowcap-=a;nowcap+=b;
maxcap=max(maxcap,nowcap);
}
cout<<maxcap<<endl;
return ;
}
C - Tram的更多相关文章
- [最短路径SPFA] POJ 1847 Tram
Tram Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 14630 Accepted: 5397 Description Tra ...
- POJ 1847 Tram (最短路)
Tram 题目链接: http://acm.hust.edu.cn/vjudge/contest/122685#problem/N Description Tram network in Zagreb ...
- poj 1847 Tram【spfa最短路】
Tram Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 12005 Accepted: 4365 Description ...
- (简单) POJ 1847 Tram,Dijkstra。
Description Tram network in Zagreb consists of a number of intersections and rails connecting some o ...
- Codeforces Round #386 (Div. 2) C. Tram
C. Tram time limit per test 1 second memory limit per test 256 megabytes input standard input output ...
- Tram
Tram 题目大意:给你一个图,这个图上有n点和边.点上有开关,开关开始指向一条道路,拨动开关可以使开关指向由开关出发的任意一条路径,读入,a,b,求,至少要拨动几次才能从a点走到b点. 注释:n&l ...
- POJ1847 Tram
Tram Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 20274 Accepted: 7553 Description ...
- poj1847 Tram(最短路dijkstra)
描述: Tram network in Zagreb consists of a number of intersections and rails connecting some of them. ...
- POJ 1847 Tram (最短路径)
POJ 1847 Tram (最短路径) Description Tram network in Zagreb consists of a number of intersections and ra ...
- POJ1847:Tram(最短路)
Tram Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 20116 Accepted: 7491 题目链接:http:/ ...
随机推荐
- leetcode刷题记录--js
leetcode刷题记录 两数之和 给定一个整数数组 nums 和一个目标值 target,请你在该数组中找出和为目标值的那 两个 整数,并返回他们的数组下标. 你可以假设每种输入只会对应一个答案.但 ...
- 转载:python 日期,季度,年份
# 这个data_matrix[:,dimen] <= thresh_val 内标会返回data_matrix当中的值符合条件的,返回为True # ret_array 中就会返回 下标为Tru ...
- Web 常用
System.Web.Hosting.HostingEnvironment.MapPath(); HttpUtility.UrlEncode();
- Luogu P1892 [BOI2003]团伙
P1892 [BOI2003]团伙 题目描述 1920年的芝加哥,出现了一群强盗.如果两个强盗遇上了,那么他们要么是朋友,要么是敌人.而且有一点是肯定的,就是: 我朋友的朋友是我的朋友: 我敌人的敌人 ...
- models中,字段参数limit_choices_to的用法
这里,在使用 ModelForm 渲染前端页面的前提下,对于 models 中的 ManyToManyField 类型字段会在 ModelForm 中被转化为 ModelMultipleChoiceF ...
- Ajax基本写法
<!DOCTYPE html> <html> <head> <meta charset="UTF-8"> <title> ...
- Selenium的安装和简单实用——PhantomJS安装
简介 Selenium是一个用于Web应用程序测试的工具. Selenium测试直接运行在浏览器中,就像真正的用户在操作一样.支持的浏览器包括IE(7, 8, 9, 10, 11),Firefox,S ...
- [转]如何有效地报告Bug
英文原文:Simon Tatham,编译:Dasn 引言 为公众写过软件的人,大概都收到过很拙劣的bug报告,例如: 在报告中说“不好用”: 所报告内容毫无意义: 在报告中用户没有提供足够的信息: 在 ...
- 洛谷 P3258 BZOJ 3631 [JLOI2014]松鼠的新家
题目描述 松鼠的新家是一棵树,前几天刚刚装修了新家,新家有n个房间,并且有n-1根树枝连接,每个房间都可以相互到达,且俩个房间之间的路线都是唯一的.天哪,他居然真的住在”树“上. 松鼠想邀请小熊维尼前 ...
- 从零搭建流媒体服务器+obs推流直播
背景介绍 本文使用的流媒体服务器的搭建是基于rtmp(Real Time Message Protocol)协议的,rtmp协议是应用层的协议,要依靠底层的传输层协议,比如tcp协议来保证信息传输的可 ...