http://codeforces.com/contest/479/problem/C

C. Exams
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 n exams. 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.

Sample test(s)
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.

解题思路:告诉你n门考试的最晚考试时间可最早考试时间,求最后一门考试的最早时间

按最晚时间升序排列,若相同则最早时间升序,贪心即可

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <stdlib.h>
 4 
 5 struct A{
 6     int s;
 7     int e;
 8 }a[];
 9 
 int cmp(const void *a, const void *b){
     struct A *c = (struct A *)a;
     struct A *d = (struct A *)b;
     if(c->s != d->s)
         return c->s - d->s;
     else return
         c->e - d->e;
 }
 
 int main(){
     int n, i, ans;
     while(scanf("%d", &n) != EOF){
         for(i = ; i < n; i++){
             scanf("%d %d", &a[i].s, &a[i].e);
         }
         qsort(a, n, sizeof(a[]), cmp);
         ans = -;
         for(i = ; i < n; i++){
             if(ans <= a[i].e){
                 ans = a[i].e;
             }
             else{
                 ans = a[i].s;
             }
         }
         printf("%d\n", ans);
     }
     return ;

38 }

Codeforces Round #274 (Div. 2)-C. Exams的更多相关文章

  1. 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 ...

  2. 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 ...

  3. Codeforces Round #377 (Div. 2) D. Exams

    Codeforces Round #377 (Div. 2) D. Exams    题意:给你n个考试科目编号1~n以及他们所需要的复习时间ai;(复习时间不一定要连续的,可以分开,只要复习够ai天 ...

  4. Codeforces Round #274 (Div. 2) 解题报告

    题目地址:http://codeforces.com/contest/479 这次自己又仅仅能做出4道题来. A题:Expression 水题. 枚举六种情况求最大值就可以. 代码例如以下: #inc ...

  5. Codeforces Round #377 (Div. 2) D. Exams(二分答案)

    D. Exams Problem Description: Vasiliy has an exam period which will continue for n days. He has to p ...

  6. Codeforces Round #377 (Div. 2) D. Exams 二分

    D. Exams time limit per test 1 second memory limit per test 256 megabytes input standard input outpu ...

  7. Codeforces Round #274 (Div. 2)

    A http://codeforces.com/contest/479/problem/A 枚举情况 #include<cstdio> #include<algorithm> ...

  8. Codeforces Round #274 (Div. 1) C. Riding in a Lift 前缀和优化dp

    C. Riding in a Lift Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/480/pr ...

  9. Codeforces Round #274 (Div. 1) B. Long Jumps 数学

    B. Long Jumps Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/480/problem/ ...

随机推荐

  1. 网络编程-http连接-GET&POST

    GetRequest package com.net.http; import java.io.BufferedReader; import java.io.IOException; import j ...

  2. 蓝桥杯T42(八数码问题)

    题目链接:http://lx.lanqiao.cn/problem.page?gpid=T42 题意:中文题诶- 思路:bfs 将没种九宫格的状态看作一个状态节点,那么只需求起始节点到目标节点的最短路 ...

  3. Python 学习 Part1

    1. 斐波那契数序列 >>> a,b=0,1 >>> a 0 >>> b 1 >>> while b<10: print( ...

  4. 用shell脚本监控MySQL主从同步

    企业面试题1:(生产实战案例):监控MySQL主从同步是否异常,如果异常,则发送短信或者邮件给管理员.提示:如果没主从同步环境,可以用下面文本放到文件里读取来模拟:阶段1:开发一个守护进程脚本每30秒 ...

  5. luoguP2408不同子串个数

    传送门 可以知道每一个子串都是后缀的前缀,那么对于第\(i\)小的后缀的贡献就可以表示为n-sa[i]+1 然而会存在重复的子串,注意height数组的定义,对于sa[i-1]和sa[i],只有hei ...

  6. eclipse中,添加JavaSE帮助文档和源码

  7. PAT甲级——1135 Is It A Red-Black Tree (30 分)

    我先在CSDN上面发表了同样的文章,见https://blog.csdn.net/weixin_44385565/article/details/88863693 排版比博客园要好一些.. 1135 ...

  8. DRF教程2-请求和响应

    Request objects REST framework中有一个Request对象,是HttpRequest的扩展,提供了新的请求解析,Request的核心功能就是request.data,它和r ...

  9. 持续集成~Jenkins里的powershell插件发布远程站点了

    通过添加powershell插件后,使用它强大的windows系统命令,就把发布好的程序包推送到具体的应用服务器了. 系统管理-插件管理-powershell 把它安装,重启jenkins,然后修改你 ...

  10. 《JavaScript设计模式》笔记之第一、二章:富有表现力的JavaScript 和 接口

    第一章 创建一个类 方法一:      var Anim = function() {           ...      };      Anim.prototype.start = functi ...