比较经典的网络流了
每条狗,如果是0就和源连v[i],否则和汇连
人也是,0和源连w[i] (+g),然后连所有指定的狗
反之亦然
考虑一下割的含义就比较显然了
#include<stdio.h>
#include<cstring>
#include<algorithm>
#define inf 2147483647
#define M 20005
#define cint const int &
using namespace std;
int tot=1,s[M],cur[M],tm[M],ts,Q[M],lev[M],v[M],dg[15],sex[M],type,ans,ED,n,m,g;
struct edge{int v,c,n;}e[M*20];
inline void push(cint u,cint v,cint c)
{
e[++tot]=(edge){v,c,s[u]};s[u]=tot;
e[++tot]=(edge){u,0,s[v]};s[v]=tot;
}
inline bool bfs()
{
memcpy(cur,s,sizeof(cur));
tm[Q[1]=1]=++ts;
for (int l=1,r=1;l<=r;l++) for (int i=s[Q[l]];i;i=e[i].n) if (e[i].c && tm[e[i].v]<ts)
tm[Q[++r]=e[i].v]=ts,lev[Q[r]]=lev[Q[l]]+1;
return tm[ED]==ts;
}
int dfs(const int &u,int c)
{
if (u==ED) return c;
int g=0,tmp;
for (int i=cur[u];i && c;cur[u]=i,i=e[i].n)
if (tm[e[i].v]==ts && lev[e[i].v]==lev[u]+1 && e[i].c && (tmp=dfs(e[i].v,min(c,e[i].c))))
g+=tmp,c-=tmp,e[i].c-=tmp,e[i^1].c+=tmp;
return g;
}
int main()
{
scanf("%d%d%d",&n,&m,&g);
for (int i=1;i<=n;i++) scanf("%d",sex+i);
for (int i=1;i<=n;i++) scanf("%d",v+i);
ED=n+m+2;
for (int i=1;i<=n;i++) if (sex[i]) push(i+1,ED,v[i]);
else push(1,i+1,v[i]);
for (int i=1,w,len;i<=m;i++)
{
scanf("%d%d%d",sex,&w,&len);ans+=w;
for (int j=1;j<=len;j++) scanf("%d",dg+j);
scanf("%d",&type);
if (sex[0]) push(n+1+i,ED,w+type*g);
else push(1,n+1+i,w+type*g);
for (int j=1;j<=len;j++)
{
if (sex[0]) push(dg[j]+1,n+1+i,inf);
else push(n+1+i,dg[j]+1,inf);
}
}
while (bfs()) for (int rt=dfs(1,inf);rt;rt=dfs(1,inf)) ans-=rt;
printf("%d",ans);
}