考慮有源匯上下界可行流：由匯向源連inf邊，那麽變成無源匯圖，按上題做法跑出可行流。此時該inf邊的流量即為原圖中該可行流的流量。因為可以假裝把加上去的那些邊的流量放回原圖。

　　此時再從原來的源向原來的匯跑最大流。超源超匯相關的邊已經流滿不會再退流，那麽可以保證滿足流量限制。求出的最大流即為原圖最大流。因為顯然原圖最大流=可行流流量+新增流量，而可行流流量等於匯到源流量，這部分在跑最大流的時候被退流並計入答案。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include
 
<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int 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<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
#define N 210
#define M 50000
#define
 
 S 0
#define T 201
#define inf 1000000000
int n,m,w,v,t=-1,p[N],degree[N],l[M],tot=0;
int cur[N],d[N],q[N],ans=0;
struct data{int to,nxt,cap,flow;
}edge[M];
void addedge(int x,int y,int z)
{
    t++;edge[t].to=y,edge[t].nxt=p[x],edge[t].cap=z,edge[t].flow=0,p[x]=t;
    t++;edge[t].to=x,edge[t].nxt=p[y],edge[t].cap=0
 
,edge[t].flow=0,p[y]=t;
}
bool bfs(int s,int t)
{
    memset(d,255,sizeof(d));d[s]=0;
    int head=0,tail=1;q[1]=s;
    do
    {
        int x=q[++head];
        for (int i=p[x];~i;i=edge[i].nxt)
        if (d[edge[i].to]==-1&&edge[i].flow<edge[i].cap)
        {
            d[edge[i].to]=d[x]+1;
            q[++tail]=edge[i].to;
        }
    }while (head<tail);
    return ~d[t];
}
int work(int k,int f,int t)
{
    if (k==t) return f;
    int used=0;
    for (int i=cur[k];~i;i=edge[i].nxt)
    if (d[k]+1==d[edge[i].to])
    {
        int w=work(edge[i].to,min(f-used,edge[i].cap-edge[i].flow),t);
        edge[i].flow+=w,edge[i^1].flow-=w;
        if (edge[i].flow<edge[i].cap) cur[k]=i;
        used+=w;if (used==f) return f;
    }
    if (used==0) d[k]=-1;
    return used;
}
void dinic(int s,int t)
{
    while (bfs(s,t))
    {
        memcpy(cur,p,sizeof(p));
        ans+=work(s,inf,t);
    }
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("loj116.in","r",stdin);
    freopen("loj116.out","w",stdout);
    const char LL[]="%I64d";
#else
    const char LL[]="%lld";
#endif
    n=read(),m=read(),w=read(),v=read();
    memset(p,255,sizeof(p));
    for (int i=1;i<=m;i++)
    {
        int x=read(),y=read(),low=read(),high=read();
        addedge(x,y,high-low);
        degree[y]+=low,degree[x]-=low;
        l[i]=low;
    }
    for (int i=1;i<=n;i++)
    if (degree[i]>0) addedge(S,i,degree[i]),tot+=degree[i];
    else if (degree[i]<0) addedge(i,T,-degree[i]);
    addedge(v,w,inf);
    dinic(S,T);
    if (ans<tot) cout<<"please go home to sleep";
    else ans=0,dinic(w,v),cout<<ans;
    return 0;
}

LOJ116 有源匯有上下界最大流（上下界網絡流）