Portal

Analysis

Establish a dot and connect a flow to each job to the amount of money that the job can make.
Each job connects to the machine it needs at a rental rate.
Each machine connects an edge with a flow rate for the purchase cost to the confluence point.
Then run the maximum flow

As to why this can be done
Sure Analogy with this question

Code

#include<bits/stdc++.h>
#define in read()
#define re register
using namespace std;
inline int read(){
	char ch;int f=1,res=0;
	while((ch=getchar())<'0'||ch>'9') if(ch=='-') f=-1;
	while(ch>='0'&&ch<='9'){
		res=(res<<1)+(res<<3)+(ch^48);
		ch=getchar();
	}
	return f==1?res:-res;
}
const int inf = 1e9;
const int N=3000,M=4e6;
int n,m;
int S,T;
int lev[N],cur[N];
int nxt[M],to[M],cap[M],ecnt=1,head[N];
inline void add(int x,int y,int w){
	nxt[++ecnt]=head[x];head[x]=ecnt;to[ecnt]=y;cap[ecnt]=w;
	nxt[++ecnt]=head[y];head[y]=ecnt;to[ecnt]=x;cap[ecnt]=0;
}
bool bfs(){
	queue<int> q;
	for(re int i=S;i<=T;++i) lev[i]=-1,cur[i]=head[i];
	q.push(S);lev[S]=0;
	while(!q.empty()){
		int u=q.front();
		q.pop();
		for(re int e=head[u];e;e=nxt[e]){
			int v=to[e];
			if(lev[v]!=-1||cap[e]<=0) continue;
			lev[v]=lev[u]+1;
			if(v==T) return 1;
			q.push(v);
		}
	}
	return 0;
}
int dinic(int u,int flow){
	if(u==T) return flow;
	int delta,res=0;
	for(re int &e=cur[u];e;e=nxt[e]){
		int v=to[e];
		if(lev[v]>lev[u]&&cap[e]){
			delta=dinic(v,min(flow-res,cap[e]));
			if(delta){
				cap[e]-=delta;cap[e^1]+=delta;
				res+=delta;if(res==flow) return res;
			}
		}
	}
	return res;
}
int main(){
	int sum=0;
	n=in;m=in;S=0;T=n+m+1;
	for(re int i=1;i<=n;++i){
		int get=in,num=in;
		add(0,i,get);sum+=get;
		for(re int j=1;j<=num;++j){
			int id=in,c=in;
			add(i,id+n,c);
		}
	}
	for(re int i=1;i<=m;++i){
		int v=in;
		add(n+i,T,v);
	}
	int maxflow=0;
	while(bfs()) maxflow+=dinic(S,inf);
	cout<<sum-maxflow;
	return 0;
}

Epilogue

stO hxy Orz
(Thanks to dalao for taking me to fly ())
Solving the problem of "simple construction"
But... I didn't think of it at first.
I want to give up. Look at the code.
But%% hxy%% encourage me to think about it.
Then ~ ~ ~
Hee-hee, same Space Flight Program That's it.
Sure to see more, think more!!!