题目:

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Student Valera is an undergraduate student at the University. His end of term exams are approaching and he is to pass exactly nexams. Valera is a smart guy, so he will be able to pass any exam he takes on his first try. Besides, he can take several exams on one day, and in any order.

According to the schedule, a student can take the exam for the i-th subject on the day number ai. However, Valera has made an arrangement with each teacher and the teacher of the i-th subject allowed him to take an exam before the schedule time on day bi(bi < ai). Thus, Valera can take an exam for the i-th subject either on day ai, or on day bi. All the teachers put the record of the exam in the student's record book on the day of the actual exam and write down the date of the mark as number ai.

Valera believes that it would be rather strange if the entries in the record book did not go in the order of non-decreasing date. Therefore Valera asks you to help him. Find the minimum possible value of the day when Valera can take the final exam if he takes exams so that all the records in his record book go in the order of non-decreasing date.

Input

The first line contains a single positive integer n (1 ≤ n ≤ 5000) — the number of exams Valera will take.

Each of the next n lines contains two positive space-separated integers ai and bi (1 ≤ bi < ai ≤ 109) — the date of the exam in the schedule and the early date of passing the i-th exam, correspondingly.

Output

Print a single integer — the minimum possible number of the day when Valera can take the last exam if he takes all the exams so that all the records in his record book go in the order of non-decreasing date.

Examples
input
3
5 2
3 1
4 2
output
2
input
3
6 1
5 2
4 3
output
6
Note

In the first sample Valera first takes an exam in the second subject on the first day (the teacher writes down the schedule date that is 3). On the next day he takes an exam in the third subject (the teacher writes down the schedule date, 4), then he takes an exam in the first subject (the teacher writes down the mark with date 5). Thus, Valera takes the last exam on the second day and the dates will go in the non-decreasing order: 3, 4, 5.

In the second sample Valera first takes an exam in the third subject on the fourth day. Then he takes an exam in the second subject on the fifth day. After that on the sixth day Valera takes an exam in the first subject.

题解:

先按a的值排序,排序完后,考虑最后一个考试要不要换,如果换了能贪心出来非降序则换,否则就不换。

 #include<algorithm>
#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
using namespace std; const int maxn=+; int n; struct Node{
int x,y;
bool operator < (const Node& tmp) const {
return x>tmp.x||(x==tmp.x)&&y>tmp.y;
}
}nds[maxn]; int main() {
// freopen("data_in.txt","r",stdin);
while (scanf("%d",&n)==&&n) {
for(int i=;i<n;i++){
scanf("%d%d",&nds[i].x,&nds[i].y);
}
sort(nds,nds+n);
int su=;
swap(nds[].x,nds[].y);
for(int i=;i<n;i++){
if(nds[i].y>nds[i-].x){
su=;
break;
}
if(nds[i].x>nds[i-].x){
swap(nds[i].x,nds[i].y);
}
}
int ans;
if(su) ans=nds[].x;
else ans=nds[].y;
printf("%d\n",ans);
}
return ;
}

CodeForces 479C Exams 贪心的更多相关文章

  1. codeforces 479C Exams 解题报告

    题目链接:http://codeforces.com/problemset/problem/479/C 题目意思:简单来说,就是有个人需要通过 n 门考试,每场考试他可以选择ai, bi 这其中一个时 ...

  2. CodeForces - 158B.Taxi (贪心)

    CodeForces - 158B.Taxi (贪心) 题意分析 首先对1234的个数分别统计,4人组的直接加上即可.然后让1和3成对处理,只有2种情况,第一种是1多,就让剩下的1和2组队处理,另外一 ...

  3. Codeforces Round #280 (Div. 2) C. Vanya and Exams 贪心

    C. Vanya and Exams Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/492/pr ...

  4. Codeforces Round #274 (Div. 1) A. Exams 贪心

    A. Exams Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/480/problem/A Des ...

  5. codeforces 480A A. Exams(贪心)

    题目链接: A. Exams time limit per test 1 second memory limit per test 256 megabytes input standard input ...

  6. Codeforces Round #377 (Div. 2) D. Exams 贪心 + 简单模拟

    http://codeforces.com/contest/732/problem/D 这题我发现很多人用二分答案,但是是不用的. 我们统计一个数值all表示要准备考试的所有日子和.+m(这些时间用来 ...

  7. Codeforces Round #481 (Div. 3) G. Petya's Exams (贪心,模拟)

    题意:你有\(n\)天的时间,这段时间中你有\(m\)长考试,\(s\)表示宣布考试的日期,\(d\)表示考试的时间,\(c\)表示需要准备时间,如果你不能准备好所有考试,输出\(-1\),否则输出你 ...

  8. Codeforces Round #274 (Div. 2) C. Exams (贪心)

    题意:给\(n\)场考试的时间,每场考试可以提前考,但是记录的是原来的考试时间,问你如何安排考试,使得考试的记录时间递增,并且最后一场考试的时间最早. 题解:因为要满足记录的考试时间递增,所以我们用结 ...

  9. codeforces 724D(贪心)

    题目链接:http://codeforces.com/contest/724/problem/D 题意:给定一个字符串和一个数字m,选取一个一个子序列s,使得对于字符串中任意长度为m的子序列都至少含有 ...

随机推荐

  1. vue+element 页面输入框--回车导致页面刷新的问题

    el-form 后面加上 @submit.native.prevent

  2. PHP 扩展 trie-tree, swoole过滤敏感词方案

    在一些app,web中评论以及一些文章会看到一些*等,除了特定的不显示外,我们会把用户输入的一些敏感字符做处理,具体显示为*还是其他字符按照业务区实现. 下面简单介绍下业务处理. 原文地址:小时刻个人 ...

  3. 详解 Python3 正则表达式(三)

    上一篇:详解 Python3 正则表达式(二) 本文翻译自:https://docs.python.org/3.4/howto/regex.html 博主对此做了一些批注和修改 ^_^ 模块级别的函数 ...

  4. python 时间time模块介绍和应用

    1.其中format_string 类型的时间和struct_time之间可以转换,timestamp时间戳可以和struct_time之间进行转化,但是时间戳和格式化时间是不能直接转换的. time ...

  5. python2.7提示编码出错

    学习python过程中经常会遇到一些问题. 比如编码出错,之前解决过.但是由于很长时间没有学习python,于是忘记解决的办法.这一次,从新开始学习又遇到了... 首先,报错提示编码出现问题: 于是通 ...

  6. python学习笔记(二):基础知识点

    python基本元素 7 // 2 3 7 % 3 1 # 取商以及余数 divmod(7,3) (2, 1) 1j*1j (-1+0j) 10/3 3.3333333333333335 '3,''1 ...

  7. WPF几个基础概念的浅显理解

    1.逻辑树与视觉树 逻辑树在结构上与xaml文件对应 视觉树更细化,拆分到控件的每个组成部分 2.依赖属性与附加属性 依赖属性:就是自己自己没有属性值,而是通过Binding从数据源获得值,就是依赖在 ...

  8. dsp6657的helloworld例程测试-第一篇

    环境搭建可以参考http://blog.sina.com.cn/s/blog_ed2e19900102xi2j.html 1. 先从mcsdk导入工程,helloworld例程 2. 提示有错误,估计 ...

  9. EmitMapper自动映射工具

             在实体与DTO之间,我们一般都需要进行映射.如果手动的来进行转换,实在是太麻烦.所以就产生了很多映射工具,比如AutoMapper,EmitMapper.而经过一些对比,EmitMa ...

  10. DOM练手讲解

    先上代码,大家贴入看一下 <body> <select id="slc" size="7"></select> <in ...