Codeforces 608 A. Saitama Destroys Hotel
1 second
256 megabytes
standard input
standard output
Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to operate an elevator in one of its other hotels. The elevator is special — it starts on the top floor, can only move down, and has infinite capacity. Floors are numbered from 0 to s and elevator initially starts on floor s at time 0.
The elevator takes exactly 1 second to move down exactly 1 floor and negligible time to pick up passengers. Genos is given a list detailing when and on which floor passengers arrive. Please determine how long in seconds it will take Genos to bring all passengers to floor 0.
The first line of input contains two integers n and s (1 ≤ n ≤ 100, 1 ≤ s ≤ 1000) — the number of passengers and the number of the top floor respectively.
The next n lines each contain two space-separated integers fi and ti (1 ≤ fi ≤ s, 1 ≤ ti ≤ 1000) — the floor and the time of arrival in seconds for the passenger number i.
Print a single integer — the minimum amount of time in seconds needed to bring all the passengers to floor 0.
3 7
2 1
3 8
5 2
11
5 10
2 77
3 33
8 21
9 12
10 64
79
In the first sample, it takes at least 11 seconds to bring all passengers to floor 0. Here is how this could be done:
1. Move to floor 5: takes 2 seconds.
2. Pick up passenger 3.
3. Move to floor 3: takes 2 seconds.
4. Wait for passenger 2 to arrive: takes 4 seconds.
5. Pick up passenger 2.
6. Go to floor 2: takes 1 second.
7. Pick up passenger 1.
8. Go to floor 0: takes 2 seconds.
This gives a total of 2 + 2 + 4 + 1 + 2 = 11 seconds.
题意就是电梯接乘客,这个电梯只能从上往下走,给的数据是在第几层有乘客,乘客到达该层的时间,乘客上电梯的时间忽略不计,要求输出花费的最短时间是多少。
我的理解就是单独把某层的时间拿出来算一下,从该层到最底层花的时间+等待乘客的时间和总的层数花的时间比较一下,找出max,然后其他层也是这样,和上一组楼层的max比较一下。其实总的想一下,就是把楼层数+等待乘客的时间找出max,然后和总的楼层数比较一下就可以了。(最近好多题都用到这种思想了,不知道这种思想叫啥,但是感觉有时候挺好用的,就是有时用这种思路写题会T,超时了耶。。。)
代码:
#include<bits/stdc++.h>
using namespace std;
int main(){
int n,m,f,t;
int maxx=-;
while(~scanf("%d%d",&n,&m)){
for(int i=;i<n;i++){
scanf("%d%d",&f,&t);
if(f+t>maxx)
maxx=max(f+t,maxx);
}
if(maxx<m)
printf("%d\n",m);
else
printf("%d\n",maxx);
}
return ;
}
Codeforces 608 A. Saitama Destroys Hotel的更多相关文章
- Codeforces Round #336 (Div. 2)A. Saitama Destroys Hotel 水题
A. Saitama Destroys Hotel 题目连接: http://www.codeforces.com/contest/608/problem/A Description Saitama ...
- Codeforces Round #336 (Div. 2) A. Saitama Destroys Hotel 模拟
A. Saitama Destroys Hotel Saitama accidentally destroyed a hotel again. To repay the hotel company ...
- Codefroces A. Saitama Destroys Hotel
A. Saitama Destroys Hotel time limit per test 1 second memory limit per test 256 megabytes input sta ...
- CodeForces - 608A-Saitama Destroys Hotel(模拟)
Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to o ...
- Codeforces 608 B. Hamming Distance Sum-前缀和
B. Hamming Distance Sum time limit per test 2 seconds memory limit per test 256 megabytes input ...
- [Codeforces #608 div2]1271D Portals
Description You play a strategic video game (yeah, we ran out of good problem legends). In this game ...
- [Codeforces #608 div2]1271C Shawarma Tent
Description The map of the capital of Berland can be viewed on the infinite coordinate plane. Each p ...
- [Codeforces #608 div2]1272B Blocks
Description There are nnn blocks arranged in a row and numbered from left to right, starting from on ...
- [Codeforces #608 div2]1271A Suits
Description A new delivery of clothing has arrived today to the clothing store. This delivery consis ...
随机推荐
- 一道前端面试题:定义一个方法将string的每个字符串间加个空格返回,调用的方式'hello world'.spacify();
偶然在群里看到了这道题:定义一个方法将string的每个字符串间加个空格返回,调用的方式'hello world'.spacify(); 这道题主要是对JavaScript对象原型的考察.
- Java-Eclipse-Jabref一条龙
Java部分: 1. 到Oracle官网下载需要版本的JDK:http://www.oracle.com/technetwork/java/javase/archive-139210.html 2. ...
- 图解WinXP局域网共享设置步骤
原文链接地址:http://blog.csdn.net/jackinzhou/article/details/8468208 第一章:共享的前提工作 1.更改不同的计算机名,设置相同的工作组! 2.我 ...
- Clevo P950笔记本加装4G模块
要补全的电路部分如下(原理图见附件) 这里经过尝试,发现左上角R217,R218不用接,3G_POWER部分不接(包括MTS3572G6.UK3018及电阻电容,3G_PWR_EN实测是3.3V,驱动 ...
- SQLMap的前世今生(Part1)
http://www.freebuf.com/sectool/77948.html 一.前言 谈到SQL注入,第一时间就会想到神器SQLMAP,SQLMap是一款用来检测与利用的SQL注入开源工具.那 ...
- POJ2492:A Bug's Life(种类并查集)
A Bug's Life Time Limit: 10000MS Memory Limit: 65536K Total Submissions: 45757 Accepted: 14757 题 ...
- bzoj4589 FWT xor版本
4589: Hard Nim Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 865 Solved: 484[Submit][Status][Disc ...
- 栈与递归的实现(Hanoi塔问题等等)
函数中有直接或间接地调用自身函数的语句,这样的函数称为递归函数.递归函数用 得好,可简化编程工作.但函数自己调用自己,有可能造成死循环.为了避免死循环,要 做到两点: (1) 降阶.递归函数虽然调用自 ...
- Python 入门学习笔记
安装和运行 官网下载安装包https://www.python.org/downloads/mac-osx/下载完直接安装即可 运行打开 terminal,输入命令 python,进入 python ...
- 「6月雅礼集训 2017 Day2」C
[题目大意] 有一棵n个点的完全二叉树,边权均为1,每个点有小鸟容量c[i] 依次来了m只小鸟,第i只小鸟初始位置在pos[i]上,问来了x只小鸟的时候,怎样安排小鸟的路线可以使得小鸟移动的边权和最小 ...