## 思路
> 问题模型：最大权闭合图
>
> 转化模型：网络最小割
这道题是网络流中一个比较重要的模型：**最大权闭合图转最大流**
建立超级源点$S$和超级汇点$T$，然后每个实验连一条从$S$到实验，流量为实验收益的边，每个仪器连一条从仪器到$T$, 流量为仪器耗费的边，然后需要的仪器就连一条从实验到仪器流量为$inf$（无穷大）的边，因为实验到仪器的边的流量为正无穷，所以最小割一定不会在上面，根据最大流最小割定理，最大流就等于最小割，我们按照以上所说建图，求出最大流，之后用实验利益的总和减去最大流，得出的就是最大净收益
最大流算法我用的是$\text{Dinic}$算法，因为这样方便输出，为什么？因为如果$d[i]$不为$0$就说明它一定用过，这样就能方便输出啦~
## 代码
```cpp
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int A = 1e5 + 11;
const int B = 1e6 + 11;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
inline int read() {
char c = getchar(); int x = 0, f = 1;
for ( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
for ( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
return x * f;
}
int m, n, cnt, opt, S, T, ans, head[A], d[A], q[A];
struct node { int from, to, nxt, val; } e[A];
inline void add(int from, int to, int val) {
e[cnt].to = to;
e[cnt].val = val;
e[cnt].nxt = head[from];
head[from] = cnt++;
}
inline bool makelevel(int s, int t) {
memset(d, 0, sizeof(d));
memset(q, 0, sizeof(q));
int l = 0, r = 0;
d[s] = 1; q[r++] = s;
while (l < r) {
int x = q[l++];
if (x == t) return true;
for (int i = head[x]; i != -1; i = e[i].nxt) {
int to = e[i].to;
if (d[to] == 0 && e[i].val > 0) {
d[to] = d[x] + 1;
q[r++] = e[i].to;
}
}
}
return false;
}
int dfs(int x, int flow, int t) {
if (x == t) return flow;
int sum = 0;
for (int i = head[x]; i != -1; i = e[i].nxt) {
int to = e[i].to;
if (d[to] == d[x] + 1 && e[i].val > 0) {
int tmp = dfs(to, min(flow - sum, e[i].val), t);
e[i].val -= tmp, e[i ^ 1].val += tmp;
sum += tmp;
if (sum == flow) return sum;
}
}
return sum;
}
int main() {
memset(head, -1, sizeof(head));
m = read(), n = read();
int S = 0, T = 555;
int w, tot = 0, x;
for (int i = 1; i <= m; i++) {
scanf("%d", &w), tot += w;
add(S, i, w), add(i, S, 0);
while (getchar() == ' ') {
scanf("%d", &x);
add(i, x + m, inf);
add(x + m, i, 0);
}
}
for (int i = 1; i <= n; i++) {
x = read();
add(i + m, T, x), add(T, i + m, 0);
}
while (makelevel(S, T)) ans += dfs(S, inf, T);
ans = tot - ans;
for (int i = 1; i <= m; i++) if (d[i]) cout << i << ' '; puts("");
for (int i = 1; i <= n; i++) if (d[i + m]) cout << i << ' '; puts("");
cout << ans << '\n';
return 0;
}
```