【bzoj3438】小M的作物 最小割

貌似应该有两种建图方法，另外一种是基于最大权闭合子图的。

一个种子和S联通表示种在A，和T联通表示重在B

源点S向第i个种子连一条流量为bi的边

第i个种子向汇点T连一条流量为ai的边

每个收益i拆成两个点i1、i2，分别表示全部种在A和全部种在B

每个收益i2向对应种子连一条流量为inf的边

源点S向每个收益i2连一条流量为c2i的边

对应种子向每个收益i1连一条流量为inf的边

每个收益i1向汇点T连一条流量为c1i的边

答案为所有收益-最小割

点数n+2m+2

边数(2n+2sumk+2m)*2

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<iostream>
#define maxn 3010
#define maxm 4100010
#define inf 1000000000

using namespace std;

int head[maxn],to[maxm],c[maxm],next[maxm],q[maxn],d[maxn];
int a[maxn],b[maxn];
int n,m,num,s,t,ans,sum;

void addedge(int x,int y,int z)
{
num++;to[num]=y;c[num]=z;next[num]=head[x];head[x]=num;
num++;to[num]=x;c[num]=0;next[num]=head[y];head[y]=num;
}

bool bfs()
{
for (int i=s;i<=t;i++) d[i]=-1;
int l=0,r=1;
q[1]=s;d[s]=0;
while (l<r)
{
int x=q[++l];
for (int p=head[x];p;p=next[p])
if (c[p] && d[to[p]]==-1)
{
d[to[p]]=d[x]+1;
q[++r]=to[p];
}
}
if (d[t]==-1) return 0; else return 1;
}

int find(int x,int low)
{
if (x==t || low==0) return low;
int totflow=0;
for (int p=head[x];p;p=next[p])
if (c[p] && d[to[p]]==d[x]+1)
{
int a=find(to[p],min(low,c[p]));
c[p]-=a;c[p^1]+=a;
totflow+=a;low-=a;
if (low==0) return totflow;
}
if (low) d[x]=-1;
return totflow;
}

void Dinic()
{
while (bfs()) ans+=find(s,inf);
}

int main()
{
scanf("%d",&n);
for (int i=1;i<=n;i++) {scanf("%d",&a[i]);sum+=a[i];}
for (int i=1;i<=n;i++) {scanf("%d",&b[i]);sum+=b[i];}
num=1;
scanf("%d",&m);
s=0;t=n+2*m+1;
for (int i=1;i<=n;i++) addedge(s,i,b[i]),addedge(i,t,a[i]);
for (int i=1;i<=m;i++)
{
int k,x,c1,c2;
scanf("%d%d%d",&k,&c1,&c2);
addedge(s,n+i*2,c2);
addedge(n+i*2-1,t,c1);
sum+=c1+c2;
for (int j=1;j<=k;j++)
{
scanf("%d",&x);
addedge(x,n+i*2-1,inf);
addedge(n+i*2,x,inf);
}
}
Dinic();
printf("%d\n",sum-ans);
return 0;
}