题意
给定 \(n\) 个工作，\(m\) 个机器。
每个工作需要若干机器获得 \(s_i\) 的奖励。
机器可以选择租和买。租只能在当前工作内使用。
Sol
考虑在最大权闭合子图上面改改。
发现直接把工作往汇点连买的权值就完事了。
Code
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <array>
#include <queue>
/* #define int long long */
#define pii pair <int, int>
using namespace std;
#ifdef ONLINE_JUDGE
#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
char buf[1 << 23], *p1 = buf, *p2 = buf, ubuf[1 << 23], *u = ubuf;
#endif
int read() {
int p = 0, flg = 1;
char c = getchar();
while (c < '0' || c > '9') {
if (c == '-') flg = -1;
c = getchar();
}
while (c >= '0' && c <= '9') {
p = p * 10 + c - '0';
c = getchar();
}
return p * flg;
}
void write(int x) {
if (x < 0) {
x = -x;
putchar('-');
}
if (x > 9) {
write(x / 10);
}
putchar(x % 10 + '0');
}
#define fi first
#define se second
const int N = 1e5 + 5, M = 2e7 + 5, inf = 2e9;
namespace G {
array <int, N> fir;
array <int, M> nex, to, cap;
int cnt = 1;
void add(int x, int y, int z) {
cnt++;
nex[cnt] = fir[x];
to[cnt] = y;
cap[cnt] = z;
fir[x] = cnt;
}
void _add(int x, int y, int z) {
add(x, y, z);
add(y, x, 0);
}
}
namespace Mfl {
array <int, N> cur, dis;
queue <int> q;
bool bfs(pii st) {
dis.fill(-1), dis[st.fi] = 0;
q.push(st.fi);
while (!q.empty()) {
int u = q.front();
q.pop();
for (int i = G::fir[u]; i; i = G::nex[i]) {
if (!G::cap[i] || ~dis[G::to[i]]) continue;
dis[G::to[i]] = dis[u] + 1, q.push(G::to[i]);
}
}
return ~dis[st.se];
}
int dfs(int x, int augcd, pii st) {
if (x == st.se) return augcd;
int augc = augcd;
for (int &i = cur[x]; i; i = G::nex[i]) {
if (!G::cap[i] || dis[G::to[i]] <= dis[x]) continue;
int flow = dfs(G::to[i], min(augc, G::cap[i]), st);
augc -= flow;
G::cap[i] -= flow, G::cap[i ^ 1] += flow;
if (!augc) break;
}
return augcd - augc;
}
int dinic(pii st) {
int ans = 0;
while (bfs(st)) {
copy(G::fir.begin(), G::fir.end(), cur.begin());
ans += dfs(st.fi, inf, st);
}
return ans;
}
}
signed main() {
int n = read(), m = read();
int ans = 0;
pii st = make_pair(n + m + 1, n + m + 2);
for (int i = 1; i <= n; i++) {
int x = read(), k = read();
ans += x;
while (k--) {
int u = read(), v = read();
G::_add(i, n + u, v);
}
G::_add(st.fi, i, x);
}
for (int i = 1; i <= m; i++)
G::_add(i + n, st.se, read());
write(ans - Mfl::dinic(st)), puts("");
return 0;
}