Ilya is a frontman of the most famous rock band on Earth. Band decided to make the most awesome music video ever for their new single. In that music video Ilya will go through Manhattan standing on the top of a huge truck and playing amazing guitar solos. And during this show residents of the island will join in singing and shaking their heads. However, there is a problem. People on some streets hate rock.
Recall that Manhattan consists of n vertical and m horizontal streets which form the grid of ( n − 1)×( m − 1) squares. Band’s producer conducted a research and realized two things. First, band’s popularity is constant on each street. Second, a popularity can be denoted as an integer from 1 to 10 9. For example, if rockers go along the street with popularity equal to 10 9 then people will greet them with a hail of applause, fireworks, laser show and boxes with... let it be an orange juice. On the other hand, if rockers go along the street with popularity equal to 1 then people will throw rotten tomatoes and eggs to the musicians. And this will not help to make the most awesome music video!
So, a route goes from the upper left corner to the bottom right corner. Let us define the route coolness as the minimal popularity over all streets in which rockers passed non-zero distance. As you have probably guessed, the musicians want to find the route with the maximal coolness. If you help them then Ilya will even give you his autograph!

Input

In the first line there are integers n and m (2 ≤ n, m ≤ 10 5), separated by space. These are the numbers of vertical and horizontal streets, respectively.
In the following n lines there are popularity values (one value on each line) on vertical streets in the order from left to right.
In the following m lines there are popularity values (one value on each line) on horizontal streets in the order from top to bottom.
It is guaranteed that all popularity values are integers from 1 to 10 9.

Output

Output a single integer which is a maximal possible route coolness.

Example

input output
2 3
4
8
2
7
3
4
4 3
12
4
12
3
21
5
16
12

解题思路:

题意:求从左上角刀右下角所经过街道的最小的权值,直接分成4种情况:不好描述,直接看代码:

#include<bits/stdc++.h>
using namespace std; int main()
{
int m,n,i,a[],b[];
cin>>m>>n;
int maxn = -,maxm = -;
for(i=;i<m;i++){
cin>>a[i];
if(a[i]>maxn&&i!=m-&&i!=)
maxn = a[i];
}
for(i=;i<n;i++){
cin>>b[i];
if(b[i]>maxm&&i!=n-&&i!=) maxm = b[i];
}
int min1 = min(b[],a[m-]);
int min2 = min(a[],b[n-]);
int min3 = min(a[],min(maxm,a[m-]));
int min4 = min(b[],min(maxn,b[n-]));
//cout<<min1<<min2<<min3<<min4<<endl;
int ans = max(max(min1,min2),max(min3,min4));
cout<<ans<<endl;
return ;
}

Hard Rock的更多相关文章

  1. ural 2069. Hard Rock

    2069. Hard Rock Time limit: 1.0 secondMemory limit: 64 MB Ilya is a frontman of the most famous rock ...

  2. POJ - 2339 Rock, Scissors, Paper

    初看题目时就发了个错误,我因为没有耐心看题而不了解题目本身的意思,找不到做题的突破口,即使看了一些题解,还是没有想到方法. 后来在去问安叔,安叔一语道破天机,问我有没有搞清题目的意思,我才恍然大悟,做 ...

  3. ROCK 聚类算法‏

    ROCK (RObust Clustering using linKs)  聚类算法‏是一种鲁棒的用于分类属性的聚类算法.该算法属于凝聚型的层次聚类算法.之所以鲁棒是因为在确认两对象(样本点/簇)之间 ...

  4. Rice Rock

    先翻译评分要点,然后一点点翻译程序实现过程 如何产生一堆岩石? rock_group = set([])#空集合,全局变量   rock_group.add(a_rock) 要画出来draw hand ...

  5. HDOJ(HDU) 2164 Rock, Paper, or Scissors?

    Problem Description Rock, Paper, Scissors is a two player game, where each player simultaneously cho ...

  6. The Rock Game

    Before the cows head home for rest and recreation, Farmer John wantsthem to get some intellectual st ...

  7. 弹指之间 -- Folk Rock

    CHAPTER 17 民谣摇滚 Folk Rock 以8Beat为主,120左右的速度最能表现此节奏特色. 示例曲目: 略

  8. 2018 ACM-ICPC 中国大学生程序设计竞赛线上赛 H题 Rock Paper Scissors Lizard Spock.(FFT字符串匹配)

    2018 ACM-ICPC 中国大学生程序设计竞赛线上赛:https://www.jisuanke.com/contest/1227 题目链接:https://nanti.jisuanke.com/t ...

  9. HDU 2164 Rock, Paper, or Scissors?

    http://acm.hdu.edu.cn/showproblem.php?pid=2164 Problem Description Rock, Paper, Scissors is a two pl ...

随机推荐

  1. 深入理解 JVM(上)

    菜鸟拙见,望请纠正(首先:推荐一本书[链接:https://pan.baidu.com/s/15I062n5LPYtRmueAAUFuFA 密码:kyo1]) 一:JVM体系概述 1:JVM是运行在操 ...

  2. RabbmitMQ-工作队列及相关概念

    工作队列-WorkQueue 实现功能: 将耗时的任务分发给多个工作者 设计思想: 避免直接去做一件资源密集型的任务,并且还得等它完成.因此将任务安排后再去做.将任务封装为一个消息,发到队列中.一个工 ...

  3. MySQL调优基础, 与hikari数据库连接池配合

    1.根据硬件配置系统参数 wait_timeout  非交互连接的最大存活时间, 10-30min max_connections   全局最大连接数 默认100 根据情况调整 back_log   ...

  4. 浅谈博弈论中的两个基本模型——Bash Game&&Nim Game

    最近在数学这一块搞了蛮多题目,已经解决了数论基础,线性代数(只有矩阵,行列式待坑),组合数学中的一些简单问题.所以接下来不可避免的对博弈论这一哲学大坑开工. 当然,由于我很菜,所以也只能从最基础最容易 ...

  5. [Oracle][Metadata]如何查找与某一个功能相关的数据字典名

    当Oracel的一个新功能出来的时候,我们可能不知道所有与此功能关联的数据字典名称,那么如何才能得到这些 meta data 的 meta data 呢? 可以通过 dicitonary 来查看: 例 ...

  6. python3通过gevent.pool限制协程并发数量

    协程虽然是轻量级的线程,但到达一定数量后,仍然会造成服务器崩溃出错.最好的方法通过限制协程并发数量来解决此类问题. server代码: #!/usr/bin/env python # -*- codi ...

  7. 使用Zabbix服务端本地邮箱账号发送报警邮件及指定报警邮件操作记录

    邮件报警有两种情况:1)Zabbix服务端只是单纯的发送报警邮件到指定邮箱,发送报警邮件的这个邮箱账号是Zabbix服务端的本地邮箱账号(例如:root@localhost.localdomain), ...

  8. Python基础系列讲解——random模块随机数的生成

    随机数参与的应用场景大家一定不会陌生,比如密码加盐时会在原密码上关联一串随机数,蒙特卡洛算法会通过随机数采样等等.Python内置的random模块提供了生成随机数的方法,使用这些方法时需要导入ran ...

  9. BugPhobia休息篇章:Beta阶段第IX次Scrum Meeting前奏

    特别说明:此次Scrum Meeting不计入正式的Scrum Meeting,因此此次工作仅为第IX次Scrum Meeting的前奏,而笔者也首次采用休息篇章作为子命题   0x01 :Scrum ...

  10. LINUX内核分析第八周学习总结

    LINUX内核分析第八周学习总结 标签(空格分隔): 20135328陈都 陈都 原创作品转载请注明出处 <Linux内核分析>MOOC课程 http://mooc.study.163.c ...