$loj\ 6045$ [雅礼集训 $2017\ Day8$] 价 网络流

正解:网络流

解题报告:

传送门$QwQ$

这题还,挺有趣的我$jio$得.

考虑依然先是照着最小割的模子建图呗,然后从意义上来分析,割一条边就相当于不吃一种减肥药/买一种药材.由已知得,买的药材数量要和吃的减肥药数量相等,所以考虑只要满足割的边数恰好等于$n$就好$QwQ$

又因为要出现割的局面至少要割$n$条(有一、、显然,,,?懒得证了$QwQ$

所以现在就要考虑割的边数一定是最少的

那不显然使流量都特别大就能实现了嘛$QwQ$

又因为是要$\sum P_{i}$取$min$,所以都变成$inf-p_{i}$就成

然后就做完辣!$over$

(感$jio$只大概说了下原理并没有详细说怎么建图?$QwQ$

就考虑减肥药和药材分别建一排点彼此之间连$inf$,药材和$T$连$inf$,$S$和减肥药连$inf-p_{i}$.

$over$

(还有就,我发现了一个双倍经验$QwQ$

#include<bits/stdc++.h>
using namespace std;
#define il inline
#define gc getchar()
#define t(i) edge[i].to
#define w(i) edge[i].wei
#define n(i) edge[i].nxt
#define ri register int
#define rb register int
#define rc register char
#define rp(i,x,y) for(ri i=x;i<=y;++i)
#define my(i,x,y) for(ri i=x;i>=y;--i)
#define e(i,x) for(ri i=head[x];~i;i=n(i))
 
const int N=+,M=+,inf=1e8;
int n,dep[N],head[N],cur[N],S,T,ed_cnt=-,as;
struct ed{int to,nxt,wei;}edge[N<<];
 
il int read()
{
    rc ch=gc;ri x=;rb y=;
    while(ch!='-' && (ch>'' || ch<''))ch=gc;
    if(ch=='-')ch=gc,y=;
    while(ch>='' && ch<='')x=(x<<)+(x<<)+(ch^''),ch=gc;
    return y?x:-x;
}
il void ad(ri x,ri y,ri z)
{edge[++ed_cnt]=(ed){x,head[y],z};head[y]=ed_cnt;edge[++ed_cnt]=(ed){y,head[x],};head[x]=ed_cnt;}
il bool bfs()
{
    queue<int>Q;Q.push(S);memset(dep,,sizeof(dep));dep[S]=;
    while(!Q.empty())
    {
        ri nw=Q.front();Q.pop();
        e(i,nw)if(w(i) && !dep[t(i)]){dep[t(i)]=dep[nw]+,Q.push(t(i));if(t(i)==T)return ;}
    }
    return ;
}
il int dfs(ri nw,ri flow)
{
    if(nw==T || !flow)return flow;ri ret=;
    for(ri &i=cur[nw];~i;i=n(i))
        if(w(i) && dep[t(i)]==dep[nw]+)
        {ri tmp=dfs(t(i),min(flow,w(i)));ret+=tmp,w(i)-=tmp;w(i^)+=tmp,flow-=tmp;}
    return ret;
}
il int dinic(){ri ret=;while(bfs()){rp(i,S,T)cur[i]=head[i];while(int d=dfs(S,inf))ret+=d;}return ret;}
 
int main()
{
    //freopen("6045.in","r",stdin);freopen("6045.out","w",stdout);
    n=read();S=;T=n<<|;memset(head,-,sizeof(head));
    rp(i,,n)ad(T,n+i,inf);
    rp(i,,n){ri t=read();while(t--){ri tmp=read();ad(tmp+n,i,inf);}}
    rp(i,,n){ri tmp=read();ad(i,S,inf-tmp);as-=inf-tmp;}
    printf("%d\n",as+dinic());
    return ;
}