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
- ajax post 请求报错Response to preflight request doesn't pass access control check: No 'Access-Control-Allow-Origin' heade
jquery ajax跨域请求,webapi webconfig配置 前台代码(放了一部分) function CheckIn(roomno) { $.ajax({ url: 'https://www ...
- python str操作
1. str.format():使用“{}”占位符格式化字符串(占位符中的索引号形式和键值对形式可以混合使用). 1 >>> string = 'python{}, django{} ...
- IDEA 创建一个普通的java项目
IntelliJ IDEA 如何创建一个普通的java项目,及创建java文件并运行 首先,确保idea软件正确安装完成,java开发工具包jdk安装完成. IntelliJ IDEA下载地址:htt ...
- css 样式 解释
字体属性:(font) 大小 {font-size: x-large;}(特大) xx-small;(极小) 一般中文用不到,只要用数值就可以,单位:PX.PD 样式 {font-style: obl ...
- 293. [NOI2000] 单词查找树——COGS
293. [NOI2000] 单词查找树 ★★ 输入文件:trie.in 输出文件:trie.out 简单对比时间限制:1 s 内存限制:128 MB 在进行文法分析的时候,通常需要检 ...
- ES解除索引只读限制
kibana dev Tools 执行:PUT _settings { "index": { "blocks": { "rea ...
- 华为USG6350防洪墙SNMP最简单功能配置
https://www.cnblogs.com/vincent-liang/p/7785089.html
- OGNL是什么
OGNL表达式是(Object-Graph Navigation Language)是对象图形化导航语言.OGNL是一个开源的项目,Struts2中默认使用OGNL表达式语言来显示数据.与Serlve ...
- string 和 vector 初探
标准库类型 string string 表示可变长的字符序列.是C++标准库类型的一部分,拥有很多优秀的性能. 定义 string 对象时如未人为初始化编译器会默认初始化为空字符串. string 对 ...
- Zend_Form 创建、校验和解析表单的基础--(手冊)
1. 创建表单对象 创建表单对象很easy:仅仅要实现 Zend_Form: <?php $form = newZend_Form; ? > 对于高级用例.须要创建 Zend_Form ...