CF memsql Start[c]UP 2.0 B

B. Distributed Join

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Piegirl was asked to implement two table join operation for distributed database system, minimizing the network traffic.

Suppose she wants to join two tables, A and B. Each of them has certain number of rows which are distributed on different number of partitions. Table A is distributed on the first cluster consisting of m partitions. Partition with index i has ai rows from A. Similarly, second cluster containing table B has n partitions, i-th one having bi rows from B.

In one network operation she can copy one row from any partition to any other partition. At the end, for each row from A and each row from B there should be a partition that has both rows. Determine the minimal number of network operations to achieve this.

Input

First line contains two integer numbers, m and n (1 ≤ m, n ≤ 105). Second line contains description of the first cluster with m space separated integers, ai (1 ≤ ai ≤ 109). Similarly, third line describes second cluster with n space separated integers, bi (1 ≤ bi ≤ 109).

Output

Print one integer — minimal number of copy operations.

Sample test(s)

input

2 2 
2 6 
3 100

output

11

input

2 3 
10 10 
1 1 1

output

6

Note

In the first example it makes sense to move all the rows to the second partition of the second cluster which is achieved in 2 + 6 + 3 = 11operations

In the second example Piegirl can copy each row from B to the both partitions of the first cluster which needs 2·3 = 6 copy operations.

简单贪心,题意不是很好理解。。

大致题意:

A有m个分支,每个分支有ai组,B有n个分支,每个分支有bi组,要将A与B合并(A的每个分支都包含B的全部,或B的每个分支都包含A的全部),使得总花费最小。

贪心策略:将a数组升序排列,b数组升序排列,求出a的和ma,b的和mb。一共就两种情况,要么将A合并到B,要要么将B合并到A,取两者的小值即可。如果是将B合并到A中,我们就将a数组遍历一下,a[i]>=mb就保留a[i],将mb合并到a[i]中,花费为mb,否则,将a[i]合并到A的其他组中,花费为a[i],注意一点如果是a的最后一项必须保留,哪怕是比mb小,也要保留下来,将mb合并进来,因为是将B合并到A,A中应该至少保留一项。

 #include<cstdio>

 #include<iostream>

 #include<cmath>

 #include<stdlib.h>

 #include<vector>

 #include<cstring>

 #include<map>

 #include<algorithm>

 #include<string.h>

 #define M(a,b) memset(a,b,sizeof(a))

 #define INF 0x3f3f3f3f

 using namespace std;

 int n,m;

 int a[],b[];

 int main()

 {

    while(scanf("%d%d",&n,&m)==)

    {

        for(int i = ;i<n;i++)

            scanf("%d",&a[i]);

        for(int i = ;i<m;i++)

            scanf("%d",&b[i]);

        sort(a,a+n);

        sort(b,b+m);

        long long sum = ;

        long long ans = ;

        if(a[n-]<b[m-])

         {

             for(int i = ;i<n;i++)

               sum+=a[i];

             for(int i = ;i<m-;i++)

             {

                 if(sum>b[i])

                 ans+=b[i];

                 else

                 ans+=sum;

             }

             ans+=sum;

         }

        else

        {

             for(int i = ;i<m;i++)

                 sum+=b[i];

             for(int i = ;i<n-;i++)

             {

                 if(sum>a[i])

                 ans+=a[i];

                 else

                 ans+=sum;

             }

             ans+=sum;

        }

         printf("%I64d\n",ans);

    }

    return ;

 }

CF memsql Start[c]UP 2.0 B的更多相关文章

  1. CF memsql Start[c]UP 2.0 A

    CF memsql Start[c]UP 2.0 A A. Golden System time limit per test 1 second memory limit per test 256 m ...

  2. 【CF MEMSQL 3.0 A. Declined Finalists】

    time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standa ...

  3. 【CF MEMSQL 3.0 E. Desk Disorder】

    time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standa ...

  4. 【CF MEMSQL 3.0 D. Third Month Insanity】

    time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standa ...

  5. 【CF MEMSQL 3.0 C. Pie Rules】

    time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standa ...

  6. 【CF MEMSQL 3.0 B. Lazy Security Guard】

    time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standa ...

  7. MemSQL Start[c]UP 2.0 - Round 1(无聊练手B题)

    http://codeforces.com/contest/452/problem/B   B. 4-point polyline time limit per test 2 seconds memo ...

  8. MemSQL Start[c]UP 2.0 - Round 2 - Online Round

    搞到凌晨4点一个没出,要gg了. A. Golden System http://codeforces.com/contest/458/problem/A #include<cstdio> ...

  9. MemSQL Start[c]UP 2.0 - Round 1

    A. Eevee http://codeforces.com/contest/452/problem/A 字符串水题 #include<cstdio> #include<cstrin ...

随机推荐

  1. 理解 OpenStack 高可用(HA)(3):Neutron 分布式虚拟路由(Neutron Distributed Virtual Routing)

    本系列会分析OpenStack 的高可用性(HA)概念和解决方案: (1)OpenStack 高可用方案概述 (2)Neutron L3 Agent HA - VRRP (虚拟路由冗余协议) (3)N ...

  2. PhotoShop算法原理解析系列 - 风格化---》查找边缘。

    之所以不写系列文章一.系列文章二这样的标题,是因为我不知道我能坚持多久.我知道我对事情的表达能力和语言的丰富性方面的天赋不高.而一段代码需要我去用心的把他从基本原理-->初步实现-->优化 ...

  3. UVALive 4426 Blast the Enemy! --求多边形重心

    题意:求一个不规则简单多边形的重心. 解法:多边形的重心就是所有三角形的重心对面积的加权平均数. 关于求多边形重心的文章: 求多边形重心 用叉积搞一搞就行了. 代码: #include <ios ...

  4. Grumpy: Go 上运行 Python!

    Google 运行数百万行Python代码用于驱动youtube.com和YouTube API的前端服务器,每秒可以提供数百万的请求! YouTube的前端运行在CPython 2.7上,所以我们投 ...

  5. C++学习笔记(2)

    本学习笔记是C++ primer plus(第六版)学习笔记.是C++学习笔记(1)的后续.复习C++基础知识的可以瞄瞄. 转载请注明出处http://www.cnblogs.com/zrtqsk/p ...

  6. AutoHotKey实现将站点添加到IE的Intranet本地站点

    最近在内部推行CRM系统,其中的CPQ组件要求必须将站点加入到"本地Intranet”才可以正常使用,但是由于使用用户比较多(超过几千人),并且每个用户的计算机水平都不一样,所以让用户手工去 ...

  7. 覆盖问题:最大覆盖问题(Maximum Covering Location Problem,MCLP)和集覆盖问题(Location Set Covering Problem,LSCP)

    集覆盖问题研究满足覆盖所有需求点顾客的前提下,服务站总的建站个数或建 设费用最小的问题.集覆盖问题最早是由 Roth和 Toregas等提出的,用于解决消防中心和救护车等的应急服务设施的选址问题,他们 ...

  8. jquery版楼层滚动特效

    <!DOCTYPE html><html> <head> <meta charset="UTF-8"> <title>楼 ...

  9. IIS不支持apk文件下载

    类型添加为:.apk MIME类型中填写apk的MIME类型“ application/vnd.android.package-archive ”

  10. C#-WinForm-Winform TextBox中只能输入数字的几种常用方法(C#)

    方法一: private void tBox_KeyPress(object sender, KeyPressEventArgs e) { if (e.KeyChar == 0x20) e.KeyCh ...