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=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_JUDGEfreopen("loj116.in","r",stdin);freopen("loj116.out","w",stdout);const char LL[]="%I64d";
#elseconst char LL[]="%lld";
#endifn=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;
}