最小費用可行流

和有源匯上下界可行流一樣建，除了加邊權外沒什麼區別。因為可以在任意節點結束，所以要對每個節點建一條到超級匯點的容量為$+\infty$邊權為0的邊。

程式碼：

#include<queue>
#include<cctype>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 305
#define M 5005
#define F inline
using namespace std;
struct edge{ int nxt,to,v,d,
 
f; }ed[M<<3];
struct fat{ int x,e; }fa[N];
int n,k=1,s,t,ss,tt,ans,h[N],d[N],r[N],a[N];
bool f[N]; queue <int> q;
F char readc(){
	static char buf[100000],*l=buf,*r=buf;
	if (l==r) r=(l=buf)+fread(buf,1,100000,stdin);
	return l==r?EOF:*l++;
}
F int _read(){
	int x=0; char ch=readc();
	while (!isdigit
 
(ch)) ch=readc();
	while (isdigit(ch)) x=(x<<3)+(x<<1)+(ch^48),ch=readc();
	return x;
}
F void add(int x,int y,int z,int w){
	if (x==0)
		x=0;
	ed[++k]=(edge){h[x],y,z,w},h[x]=k;
	ed[++k]=(edge){h[y],x,0,-w},h[y]=k;
}
F int spfa(){
	for (int i=ss;i<=tt;i++) d[i]=1e9,f[i]=false;
	while (!
 
q.empty()) q.pop();
	for (q.push(ss),d[ss]=0,r[ss]=1e9;!q.empty();){
		int x=q.front(); q.pop(),f[x]=false;
		for (int i=h[x],v;i;i=ed[i].nxt)
			if (ed[i].v>ed[i].f&&d[x]+ed[i].d<d[v=ed[i].to]){
				d[v]=d[x]+ed[i].d,fa[v].e=i,fa[v].x=x;
				r[v]=min(r[x],ed[i].v-ed[i].f);
				if (!f[v]) q.push(v),f[v]=true;
			}
	}
	return d[tt]==1e9?0:r[tt];
}
int main(){
	n=_read(),s=1,t=n+1,tt=n+2;
	for (int i=1,m;i<=n;i++){ 
		m=_read(),a[i]-=m,add(i,t,1e9,0);
		for (int v,d;m;m--)
			v=_read(),d=_read(),a[v]++,ans+=d,add(i,v,1e9,d);
	}
	for (int i=1;i<=n;i++)
		if (a[i]>0) add(ss,i,a[i],0);
		else if (a[i]<0) add(i,tt,-a[i],0);
	add(t,s,1e9,0);
	for (int sum;sum=spfa();ans+=d[tt]*sum)
		for (int x=tt,e;x!=ss;x=fa[x].x)
			ed[e=fa[x].e].f+=sum,ed[e^1].f-=sum;
	return printf("%d\n",ans),0;
}