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;
}