如果没有租用机器就是一个裸的最大权闭合子图。现在有了租用机器应该怎么办呢？
单独拆点是不行的，因为会和直接买下的情况脱离关系，租借是和连边直接相关的，那么就可以考虑在边上下功夫。我们把从工程到机器的连边从\(INF\)变成租用费用，这样再跑最大权闭合子图，就可以同时考虑租用费用了。如果租用费用过高就会割断（选用）购买机器的费用，反之亦然。
#include <bits/stdc++.h>
using namespace std;
const int N = 2410;
const int M = 4000010;
const int INF = 0x3f3f3f3f;
struct Graph {
int cnt, head[N];
struct edge {
int nxt, to, f;
}e[M];
void Init () {
cnt = -1;
memset (head, -1, sizeof (head));
}
void add_len (int u, int v, int f) {
e[++cnt] = (edge) {head[u], v, f}; head[u] = cnt;
e[++cnt] = (edge) {head[v], u, 0}; head[v] = cnt;
}
queue <int> q;
int cur[N], deep[N];
bool bfs (int s, int t) {
memcpy (cur, head, sizeof (head));
memset (deep, 0x3f, sizeof (deep));
deep[s] = 0; q.push (s);
while (!q.empty ()) {
int u = q.front (); q.pop ();
for (int i = head[u]; ~i; i = e[i].nxt) {
int v = e[i].to;
if (deep[v] == INF && e[i].f) {
deep[v] = deep[u] + 1;
q.push (v);
}
}
}
return deep[t] != INF;
}
int dfs (int u, int t, int lim) {
if (u == t || !lim) return lim;
int tmp = 0, flow = 0;
for (int &i = cur[u]; ~i; i = e[i].nxt) {
int v = e[i].to;
if (deep[v] == deep[u] + 1) {
tmp = dfs (v, t, min (lim, e[i].f));
lim -= tmp;
flow += tmp;
e[i ^ 0].f -= tmp;
e[i ^ 1].f += tmp;
if (!lim) break;
}
}
return flow;
}
int Dinic (int s, int t) {
int min_cut = 0;
while (bfs (s, t)) {
min_cut += dfs (s, t, INF);
}
return min_cut;
}
}G;
int n, m;
int A (int x) {return n * 0 + x;}
int B (int x) {return n * 1 + x;}
int main () {
//freopen ("data.in", "r", stdin);
cin >> n >> m; G.Init ();
int s = N - 1, t = N - 2, ans = 0;
for (int i = 1; i <= n; ++i) {
static int w, k, p, r;
cin >> w >> k; ans += w;
G.add_len (s, A (i), w); //get_val = w
for (int j = 1; j <= k; ++j) {
cin >> p >> r;
G.add_len (A (i), B (p), r); //rent cost = r
}
}
for (int i = 1; i <= m; ++i) {
static int w; cin >> w;
G.add_len (B (i), t, w);
}
cout << ans - G.Dinic (s, t) << endl;
}