Description:
In order to increase the happiness index of people's lives, King Y has decided to develop the manufacturing industry vigorously. There are total n kinds of products that King can choose to produce, different products can improve the happiness index of poeple's lives in different degrees, of course, the production of goods needs raw materials, different products need different ore or other products as raw materials. There are total m mines, and each mine can exploit different ore, Therefore, there are m types of ores, the cost of each mining for each mine is different, king Y want to maximize the income, the calculation method of income is:∑increased happiness index - ∑mining costs.
If you choose to exploit a mine, there will be an unlimited number of this kind of ore. What's more, if you produce one product, the happiness index will definitely increase, no matter how many you produce.
Input:
The first line of the input has an integer T(1<=T<=50), which represents the number of test cases.
In each test case, the first line of the input contains two integers n(1<=n<=200)--the number of the products and m(1<=m<=200)--the number of mines. The second line contains n integers, val[i] indicates the happiness index that number i product can increase. The third line contains m integers, cost[i] indicates the mining cost of number i mine. The next n lines, each line describes the type of raw material needed for the number i product, in each line, the first two integers n1(1<=n1<=m)--the number of ores that this product needs, n2(1<=n2<=n)--the number of products that this product needs, the next n1 + n2 integers indicate the id of ore and product that this product needs. it guarantees that ∑n1+∑n2<=2000.
Output:
Each test case output an integer that indicates the maximum value ∑val[i]-∑cost[i].
忽略每行输出的末尾多余空格
样例输入
2
3 3
600 200 400
100 200 300
1 2 1 2 3
1 0 2
1 0 3
3 4
600 400 200
100 200 300 1000
2 1 1 2 3
1 0 1
1 0 1
样例输出
600
900
思路：
最大权闭合子图，闭合图的概念为：一些点的集合，集合中的点的出边都在这个集合内。如果这些点有权值，则可以找最大权闭合子图
这种模型明显适用于点之间有依赖关系（先后顺序？）的图（DAG等等）
明显是权中有正有负所以我们源点连正权的点，容量为正权，汇点连负权点，容量为父权的绝对值，然后有依赖关系的连边，容量为INF
然后答案是正权之和-最小割（最大流）
证明请看https://wenku.baidu.com/view/986baf00b52acfc789ebc9a9.html
accode
#include<bits/stdc++.h>
#define LL long long
#define INF 0x3f3f3f3f3f3f3f3f
using namespace std;
const int MX = 500+4;
const int MS = 500*500+32;
string s[31];
int n,m;
int cost1[MX];
int cost2[MX];
template<class T>
struct Max_Flow
{
int n;
int Q[MX],sign;
int head[MX],level[MX],cur[MX],pre[MX];
int nxt[MS],pnt[MS],E;
T cap[MS];
void init(int n)
{
E = 0;
this->n = n+1;
fill(head,head+this->n,-1);
}
void add(int from, int to,T c,T rw = 0){
pnt[E] = to;
cap[E] = c;
nxt[E] = head[from];
head[from] = E++;
pnt[E] = from;
cap[E] = rw;
nxt[E] =head[to];
head[to] = E++;
}
bool BFS(int s,int t)
{
sign = t;
std::fill(level,level+n,-1);
int *front = Q;
int *tail = Q;
*tail++ = t;
level[t] = 0;
while(front < tail &&level[s] == -1){
int u = *front++;
for(int e = head[u];e!=-1;e = nxt[e]){
if(cap[e^1] > 0&&level[pnt[e]]<0){
level[pnt[e]] = level[u]+1;
*tail++ = pnt[e];
}
}
}
return level[s] != -1;
}
void Push(int t,T &flow){
T mi =INF;
int p = pre[t];
for(int p = pre[t];p!=-1;p = pre[pnt[p^1]]){
mi = std::min(mi,cap[p]);
}
for(int p = pre[t];p!=-1;p = pre[pnt[p^1]]){
cap[p] -= mi;
if(!cap[p]){
sign = pnt[p^1];
}
cap[p^1] += mi;
}
flow += mi;
}
void DFS(int u,int t, T &flow){
if(u==t){
Push(t,flow);
return ;
}
for(int &e = cur[u];e != -1;e = nxt[e]){
if(cap[e]>0&&level[u]-1==level[pnt[e]]){
pre[pnt[e]] = e;
DFS(pnt[e],t,flow);
if(level[sign] >level[u]){
return;
}
sign = t;
}
}
}
T Dinic(int s,int t){
pre[s] = -1;
T flow = 0;
while(BFS(s,t)){
std::copy(head,head+n,cur);
DFS(s,t,flow);
}
return flow;
}
};
Max_Flow<LL>F;
int main()
{
int t;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&m);
F.init(n+m+2);
LL sum = 0;
for(int i = 1;i<=n;i++){
scanf("%d",&cost1[i]);
sum += cost1[i];
F.add(0,i,cost1[i]);
}
for(int i = 1;i<=m;i++){
scanf("%d",&cost2[i]);
F.add(i+n,n+m+2,cost2[i]);
}
for(int i = 1;i<=n;i++){
int c1,c2;
scanf("%d%d",&c1,&c2);
for(int j = 0;j<c1;j++){
int x;
scanf("%d",&x);
F.add(i,x+n,INF);
}
for(int j = 0;j<c2;j++){
int x;
scanf("%d",&x);
F.add(i,x,INF);
}
}
LL ans = F.Dinic(0,n+m+2);
// cout<<ans<<endl;
printf("%ld\n",sum-ans);
}
}