loj6045 价

题目链接

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;
}