LOJ116 有源汇有上下界最大流（上下界网络流）

　　考虑有源汇上下界可行流：由汇向源连inf边，那么变成无源汇图，按上题做法跑出可行流。此时该inf边的流量即为原图中该可行流的流量。因为可以假装把加上去的那些边的流量放回原图。

　　此时再从原来的源向原来的汇跑最大流。超源超汇相关的边已经流满不会再退流，则下界可以满足，并且在此基础上增广是可以保证原图的流量平衡的。求出的最大流即为原图最大流。因为显然原图最大流=可行流流量+原图新增流量，而可行流流量等于汇到源流量，这部分在跑最大流的时候被退流并计入答案。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),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=-,p[N],degree[N],l[M],tot=;
int cur[N],d[N],q[N],ans=;
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=,p[x]=t;
    t++;edge[t].to=x,edge[t].nxt=p[y],edge[t].cap=,edge[t].flow=,p[y]=t;
}
bool bfs(int s,int t)
{
    memset(d,,sizeof(d));d[s]=;
    int head=,tail=;q[]=s;
    do
    {
        int x=q[++head];
        for (int i=p[x];~i;i=edge[i].nxt)
        if (d[edge[i].to]==-&&edge[i].flow<edge[i].cap)
        {
            d[edge[i].to]=d[x]+;
            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=;
    for (int i=cur[k];~i;i=edge[i].nxt)
    if (d[k]+==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^].flow-=w;
        if (edge[i].flow<edge[i].cap) cur[k]=i;
        used+=w;if (used==f) return f;
    }
    if (used==) d[k]=-;
    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,,sizeof(p));
    for (int i=;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=;i<=n;i++)
    if (degree[i]>) addedge(S,i,degree[i]),tot+=degree[i];
    else if (degree[i]<) addedge(i,T,-degree[i]);
    addedge(v,w,inf);
    dinic(S,T);
    if (ans<tot) cout<<"please go home to sleep";
    else ans=,dinic(w,v),cout<<ans;
    return ;
}