Xiao Ming is an expert in computer science and technology, so he can get a lot of projects every month. The projects always bring him a lot of money, now he is thinking how to earn money as more as possible.

Every month he can get m projects, and each project Ai will bring him Xi yuan. Although Xiao Ming is an expert, he still needs to hire some other guys to help him. Of course, the employees are not as good as Xiao Ming, for they are just good at some single aspect. So, they should work together to finish one project. There is a list shows the salary of m employees, who are labeled from 0 to m-1. Xiao Ming only hires employees, in that list, and he knows who will be needed by each project.If one employee is hired, he can join in several projects.

Input

The first line is an integer c shows the number of cases. For each case, the first line has two numbers m,n(m,n <=100), denoting that there is m projects and n employees on the list.The second line has m integers, which are seperated by a single blank, the ith number Ximeans the project Ai will bring Xiao Ming Xi yuan. Xi is less the 10000. The third line has n integers, which are seperated by a single blank, the ith number Yimeans the employee Bi will cost Xiao Ming Yi yuan. And the next m lines will show which part of the employees will be needed by each project. Line i is a list of the employees, who are needed by project Ai. In each line, first a number Zi shows the number of employees needed by this project. And Zi labels of the emloyees follows, which are still seperated by a sigle blank.

Output

You should output a single integer shows the maximun money Xiao Ming can earn in a single month. The money he can earn is equall to the money he can totally get minus the money he totally cost. You should not leave any extra blanks at the end of each line.

Sample Input

1
3 5
30 40 43
55 17 23 22 11
3 0 1 2
3 1 2 3
2 2 1

Sample Output

21

Hint

If Xiao Ming can do less project to earn more money, he will certainly do that.

题解:

最大权闭合子图水题,每一个工作任务有一条边 容量为输入的利润,每一个员工到汇点有一条边 容量为雇佣费,

每一个工作任务到需要的员工有一条边,容量为INF.

答案就是利润和-最小割

 #include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cstdlib>
using namespace std;
const int N=,INF=;
int gi(){
int str=;char ch=getchar();
while(ch>''||ch<'')ch=getchar();
while(ch>='' && ch<='')str=str*+ch-'',ch=getchar();
return str;
}
int n,m,s=,T,ans=;
int num=,head[N],q[N],dep[N];
struct Lin{
int next,to,dis;
}a[N*N];
void init(int x,int y,int dis){
a[++num].next=head[x];
a[num].to=y;
a[num].dis=dis;
head[x]=num;
a[++num].next=head[y];
a[num].to=x;
a[num].dis=;
head[y]=num;
}
bool bfs()
{
memset(dep,,sizeof(dep));
q[]=s;dep[s]=;int t=,sum=,u,x;
while(t!=sum)
{
x=q[++t];
for(int i=head[x];i;i=a[i].next){
u=a[i].to;
if(dep[u]||a[i].dis<=)continue;
dep[u]=dep[x]+;q[++sum]=u;
}
}
return dep[T];
}
int dfs(int x,int flow)
{
if(x==T || !flow)return flow;
int u,tmp,sum=;
for(int i=head[x];i;i=a[i].next){
u=a[i].to;
if(a[i].dis<= || dep[u]!=dep[x]+)continue;
tmp=dfs(u,min(flow,a[i].dis));
a[i].dis-=tmp;a[i^].dis+=tmp;
sum+=tmp;flow-=tmp;
}
return sum;
}
int maxflow(){
int tot=,tmp;
while(bfs()){
tmp=dfs(s,INF);
while(tmp)tot+=tmp,tmp=dfs(s,INF);
}
return tot;
}
void work()
{
int x,k;
m=gi();n=gi();
T=n+m+;
for(int i=;i<=m;i++){
x=gi();init(s,i,x);
ans+=x;
}
for(int i=;i<=n;i++){
x=gi();init(i+m,T,x);
}
for(int i=;i<=m;i++){
k=gi();
while(k--){
x=gi();init(i,x+m+,INF);
}
}
printf("%d\n",ans-maxflow());
}
void Clear(){
memset(head,,sizeof(head));
num=;ans=;
}
int main()
{
int TT=gi();
while(TT--){
work();
Clear();
}
return ;
}

[HOJ2634] How to earn more 最大权闭合子图的更多相关文章

  1. Human life FZU - 2295 最大权闭合子图(第一次遇到被教育了)

    Xzz is playing a MMORPG "human life". In this game, there are N different skills. Some ski ...

  2. BZOJ1565 [NOI2009]植物大战僵尸(拓扑排序 + 最大权闭合子图)

    题目 Source http://www.lydsy.com/JudgeOnline/problem.php?id=1565 Description Input Output 仅包含一个整数,表示可以 ...

  3. HDU 3879 Base Station(最大权闭合子图)

    经典例题,好像说可以转化成maxflow(n,n+m),暂时只可以勉强理解maxflow(n+m,n+m)的做法. 题意:输入n个点,m条边的无向图.点权为负,边权为正,点权为代价,边权为获益,输出最 ...

  4. [BZOJ 1497][NOI 2006]最大获利(最大权闭合子图)

    题目:http://www.lydsy.com:808/JudgeOnline/problem.php?id=1497 分析: 这是在有向图中的问题,且边依赖于点,有向图中存在点.边之间的依赖关系可以 ...

  5. HDU4971 A simple brute force problem.(强连通分量缩点 + 最大权闭合子图)

    题目 Source http://acm.hdu.edu.cn/showproblem.php?pid=4971 Description There's a company with several ...

  6. HDU5855 Less Time, More profit(最大权闭合子图)

    题目 Source http://acm.hdu.edu.cn/showproblem.php?pid=5855 Description The city planners plan to build ...

  7. HDU5772 String problem(最大权闭合子图)

    题目..说了很多东西 官方题解是这么说的: 首先将点分为3类 第一类:Pij 表示第i个点和第j个点组合的点,那么Pij的权值等于w[i][j]+w[j][i](表示得分) 第二类:原串中的n个点每个 ...

  8. SCU3109 Space flight(最大权闭合子图)

    嗯,裸的最大权闭合子图. #include<cstdio> #include<cstring> #include<queue> #include<algori ...

  9. hiho 第119周 最大权闭合子图

    描述 周末,小Hi和小Ho所在的班级决定举行一些班级建设活动. 根据周内的调查结果,小Hi和小Ho一共列出了N项不同的活动(编号1..N),第i项活动能够产生a[i]的活跃值. 班级一共有M名学生(编 ...

随机推荐

  1. js计时功能

    //个位秒加 function time4jia() { //分钟60为上限 所有加停止 if (sz(a('time1').innerHTML) == 6) { return; } var m4 = ...

  2. 20162330 实验二 《Java面向对象程序设计》 实验报告

    2016-2017-2 实验报告目录: 1 2 3 4 5 20162330 实验二 <Java面向对象程序设计> 实验报告 课程名称:<程序设计与数据结构> 学生班级:162 ...

  3. 学生ID查询

    var http = require("http"); var server = http.createServer(function(req,res){ //得到url var ...

  4. 400多个开源项目以及43个优秀的Swift开源项目-Swift编程语言资料大合集

    Swift 基于C和Objective-C,是供iOS和OS X应用编程的全新语言,更加高效.现代.安全,可以提升应用性能,同时降低开发难度. Swift仍然处于beta测试的阶段,会在iOS 8发布 ...

  5. style scoped

    scoped: 只在父div和其内容内生效,

  6. PHP截取日期

    date( 'Y-m-d ',strtotime('2017-10-9 12:23:35')) 通过时间格式,获取的是2017-10-9

  7. 在bootstrap中让竖向排列的输入框水平排列

    在bootstrap中可以使用自带的样式标记来控制样式,但是同时可以利用最原始的css样式来解决达到需求 如下所示可以看出来两个inline-block就可以使得两个水平排列 block和inline ...

  8. 20170222==(MODBUS读取多个寄存器)

    MODBUS读取多个寄存器(功能码04) 为了简单我这里只用4个寄存器,当让你也可以用125个寄存器,但是最多也只能用125个寄存器的.每个寄存器有上面的表知道为一个字的大小即2个字节或者叫16比特位 ...

  9. Linq SelectMany 交叉连接

    class Student1 { public int Score { get; set; } public Student1(int score) { this.Score = score; } } ...

  10. mybatis的mapper接口代理使用的三个规范

    1.什么是mapper代理接口方式? MyBatis之mapper代理方式.mapper代理使用的是JDK的动态代理策略 2.使用mapper代理方式有什么好处 使用这种方式可以不用写接口的实现类,免 ...