思路

从源点(s)向每种药连一条边权为(-p+inf)的边。从每种药向他所需要的药材连一条边权为(inf)的边。从每种药材向汇点(t)连一条边权为(inf)的边。
(inf>inf)
用最小割减去源点连向药材的边权之和。

代码

#include<cstdio> #include<iostream> #include<cstdlib> #include<cstring> #include<queue> #include<cmath> #include<ctime> #include<bitset> using namespace std; typedef long long ll; const int n = 100010,inf = 5e6,inf = 1e9; ll read() {     ll x=0,f=1;char c=getchar();     while(c<'0'||c>'9') {         if(c=='-') f=-1;         c=getchar();     }     while(c>='0'&&c<='9') {         x=x*10+c-'0';         c=getchar();     }     return x*f; } int n; int s,t; struct node {     int v,nxt,w; }e[n << 1]; int head[n],ejs = 1; void add(int u,int v,int w) {     e[++ejs].v = v;e[ejs].nxt = head[u];head[u] = ejs;e[ejs].w = w;     e[++ejs].v = u;e[ejs].nxt = head[v];head[v] = ejs;e[ejs].w = 0; } int dep[n]; queue<int>q; int bfs() {     memset(dep,0,sizeof(dep));     while(!q.empty()) q.pop();     q.push(s);dep[s] = 1;     while(!q.empty()) {         int u = q.front();q.pop();         for(int i = head[u];i;i = e[i].nxt) {             int v = e[i].v;             if(dep[v] || e[i].w <= 0) continue;             dep[v] = dep[u] + 1;             q.push(v);             if(v == t) return 1;         }     }     return 0; } int dfs(int u,int now) {     if(u == t) return now;     int ret = 0;     for(int i = head[u];i;i = e[i].nxt) {         int v = e[i].v;         if(e[i].w > 0 && dep[v] == dep[u] + 1) {             int k = dfs(v,min(now - ret,e[i].w));             ret += k;             e[i].w -= k;             e[i ^ 1].w += k;             if(ret == now) return ret;         }     }     return ret; } ll dinic() {     ll ans = 0;     while(bfs())         ans += dfs(s,inf);     return ans; } int main() {     n = read();     ll tot = 0;     s = n * 2 + 1,t = s + 1;     for(int i = 1;i <= n;++i) {         int k = read();         for(int j = 1;j <= k;++j) {             int t = read();             add(i,t + n,inf);         }     }     for(int i = 1;i <= n;++i) {         int w = read();         tot += inf - w;         add(s,i,inf - w);         add(i + n,t,inf);     }     cout<<dinic() - tot;     return 0; }