题目链接:http://codeforces.com/problemset/problem/589/F

A gourmet came into the banquet hall, where the cooks suggested n dishes for guests. The gourmet knows the schedule: when each of the dishes will be served.

For i-th of the dishes he knows two integer moments in time ai and bi (in seconds from the beginning of the banquet) — when the cooks will bring the i-th dish into the hall and when they will carry it out (ai < bi). For example, if ai = 10 and bi = 11, then the i-th dish is available for eating during one second.

The dishes come in very large quantities, so it is guaranteed that as long as the dish is available for eating (i. e. while it is in the hall) it cannot run out.

The gourmet wants to try each of the n dishes and not to offend any of the cooks. Because of that the gourmet wants to eat each of the dishes for the same amount of time. During eating the gourmet can instantly switch between the dishes. Switching between dishes is allowed for him only at integer moments in time. The gourmet can eat no more than one dish simultaneously. It is allowed to return to a dish after eating any other dishes.

The gourmet wants to eat as long as possible on the banquet without violating any conditions described above. Can you help him and find out the maximum total time he can eat the dishes on the banquet?

Input

The first line of input contains an integer n (1 ≤ n ≤ 100) — the number of dishes on the banquet.

The following n lines contain information about availability of the dishes. The i-th line contains two integers ai and bi (0 ≤ ai < bi ≤ 10000) — the moments in time when the i-th dish becomes available for eating and when the i-th dish is taken away from the hall.

Output

Output should contain the only integer — the maximum total time the gourmet can eat the dishes on the banquet.

The gourmet can instantly switch between the dishes but only at integer moments in time. It is allowed to return to a dish after eating any other dishes. Also in every moment in time he can eat no more than one dish.

Examples

Input
3
2 4
1 5
6 9
Output
6
Input
3
1 2
1 2
1 2
Output
0

Note

In the first example the gourmet eats the second dish for one second (from the moment in time 1 to the moment in time 2), then he eats the first dish for two seconds (from 2 to 4), then he returns to the second dish for one second (from 4 to 5). After that he eats the third dish for two seconds (from 6 to 8).

In the second example the gourmet cannot eat each dish for at least one second because there are three dishes but they are available for only one second (from 1 to 2).

题目大意:输入n,代表有n种菜,下面n行代表每种菜上来的时间和下去的时间,要求你每道菜吃的时间一样多,问你最多可以吃多少多久,每一秒只能吃一道菜

思路:并不是自己的思路,自己不知道怎么贪心,大概猜出来是二分求值。。。 贪心的思想是按照每一道菜的端下去的时间从小到大排序,然后从前往后选择就行。。。这是为何呢?

因为我们要做的就是让选择的这道菜对其他菜影响最可能的少,然后我们按照端下去的时间排序,证明我们当前吃的菜端下去的时间是在其他菜端下去之前的,那么我们从前往后找时间是不是对其他菜影响最小呢?  当然是,接下来就看代码了

#include<iostream>
#include<string.h>
#include<map>
#include<cstdio>
#include<cstring>
#include<stdio.h>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<set>
#include<queue>
typedef long long ll;
using namespace std;
const ll mod=1e9+;
const int maxn=1e2+;
const int maxk=1e4+;
const int maxx=1e4+;
const ll maxe=+;
#define INF 0x3f3f3f3f3f3f
#define Lson l,mid,rt<<1
#define Rson mid+1,r,rt<<1|1]
bool vis[maxk];
int n;
struct dish
{
int be,en;
}d[maxn];
bool cmp(const dish a,const dish b)
{
return a.en<b.en;
}
bool solve(int mid)
{
for(int i=;i<n;i++)
{
int sum=;
for(int j=d[i].be;j<d[i].en;j++)
{
if(!vis[j])
{
vis[j]=true;
sum++;
}
if(sum>=mid) break;
}
if(sum<mid)
return false;
}
return true;
}
int main()
{
memset(vis,false,sizeof(vis));
int mi=maxk;
cin>>n;
for(int i=;i<n;i++)
{
cin>>d[i].be>>d[i].en;
mi=min(abs(d[i].en-d[i].be),mi);
}
sort(d,d+n,cmp);
int l=,r=mi,mid=mi,ans=mid;
while(l<=r)
{
memset(vis,false,sizeof(vis));
if(solve(mid))
{
l=mid+;
ans=mid;
}
else
r=mid-;
mid=(l+r)/;
}
cout<<ans*n<<endl;
return ;
}

F. Gourmet and Banquet(贪心加二分求值)的更多相关文章

  1. Luogu2869 [USACO07DEC]美食的食草动物Gourmet Grazers (贪心,二分,数据结构优化)

    贪心 考场上因无优化与卡常T掉的\(n \log(n)\) //#include <iostream> #include <cstdio> #include <cstri ...

  2. 【CodeForces 589F】Gourmet and Banquet(二分+贪心或网络流)

    F. Gourmet and Banquet time limit per test 2 seconds memory limit per test 512 megabytes input stand ...

  3. BZOJ3527 推出卷积公式FFT求值

    BZOJ3527 推出卷积公式FFT求值 传送门:https://www.lydsy.com/JudgeOnline/problem.php?id=3527 题意: \(F_{j}=\sum_{i&l ...

  4. hdu 3641 数论 二分求符合条件的最小值数学杂题

    http://acm.hdu.edu.cn/showproblem.php?pid=3641 学到: 1.二分求符合条件的最小值 /*================================= ...

  5. 九度OJ 1085 求root(N, k) -- 二分求幂及快速幂取模

    题目地址:http://ac.jobdu.com/problem.php?pid=1085 题目描述: N<k时,root(N,k) = N,否则,root(N,k) = root(N',k). ...

  6. hdu5256 二分求LIS+思维

    解题的思路很巧,为了让每个数之间都留出对应的上升空间,使a[i]=a[i]-i,然后再求LIS 另外二分求LIS是比较快的 #include<bits/stdc++.h> #define ...

  7. 二分求幂/快速幂取模运算——root(N,k)

    二分求幂 int getMi(int a,int b) { ; ) { //当二进制位k位为1时,需要累乘a的2^k次方,然后用ans保存 == ) { ans *= a; } a *= a; b / ...

  8. 二分求幂,快速求解a的b次幂

    一个引子 如何求得a的b次幂呢,那还不简单,一个for循环就可以实现! void main(void) { int a, b; ; cin >> a >> b; ; i < ...

  9. nyoj 409——郁闷的C小加(三)——————【中缀式化前缀后缀并求值】

    郁闷的C小加(三) 时间限制:1000 ms  |  内存限制:65535 KB 难度:4   描述 聪明的你帮助C小加解决了中缀表达式到后缀表达式的转换(详情请参考“郁闷的C小加(一)”),C小加很 ...

随机推荐

  1. 洛谷【P2005】A/B Problem II

    题目传送门:https://www.luogu.org/problemnew/show/P2005 高精除低精:https://www.cnblogs.com/AKMer/p/9724556.html ...

  2. tx1 高温不启动问题解决办法

    执行: vim /boot/extlinux/extlinux.conf 修改为下面红框标注的样式: 拷贝此文件下所有内容到/boot目录下. 检查是否成功:

  3. swift-get-nodes简单使用

    在参考http://blog.csdn.net/cywosp/article/details/12850645文章对对象的具体物理磁盘位置进行查找时,发现两个问题: 1. 在使用swift+keyst ...

  4. modbus读输入状态与读线圈状态的区别?

    01 读线圈状态 描述 读从机离散量输出口的 ON/OFF 状态,不支持广播.附录B列出由不同控制器型号支持最大的参数清单. 查询 查询信息规定了要读的起始线圈和线圈量,线圈的起始地址为零,1-16个 ...

  5. nginx做代理部署WordPress

    实验环境:CentOS7 服务器172.16.252.142做Nginx代理服务器: [root@conf.d localhost]#iptables -F [root@conf.d localhos ...

  6. [.net] 无法创建虚拟目录。已将URL“XXX”映射到IIS Express网站上的一个不同的文件夹

    工作时,在修改项目属性,Web中服务器时,出现了下面的错误: 各种折腾后,找到下面的解决方法: 1.找到项目在本地的目录,目录下有当前项目的项目文件,文件名以.csproj为后缀名. 2.用文本编辑软 ...

  7. eclipse tomcat 无法加载导入的web项目,There are no resources that can be added or removed from the server. .

    应该是项目自己的setting文件夹下的描述信息和.project文件的描述信息,不能适用于这个eclipse和tomcat. 解决方法: 1,找相同类型的工程(tomcat能引用的)2,把新建项目里 ...

  8. Jquery隐藏相同name的div

    $("div:[name=divName]").hide(); divName(自己div的Name)

  9. VIsual Studio 2010 常用快捷键

    1.Ctrl+S   保存 2.Ctrl+F: 查找 3.Ctrl+H: 替换 4.Ctrl+E,S: 查看空白 5.Ctrl+K+C: 注释选定内容 6.Ctrl+K+U: 取消选定注释内容 7.C ...

  10. Ubuntu 解决:当执行`sudo apt-get update`命令时 出现的 “apt-get 404 Not Found Package Repository Errors” 问题

    Ubuntu 解决:当执行sudo apt-get update或者sudo apt-get install命令是出现的 "apt-get 404 Not Found Package Rep ...