假设初始流为每条边的下界。但这样可能流量会不守恒，我们需要在上面加上一个附加流使流量守恒。只要让每个点开始的出/入流量与原初始流相等就可以求出附加流了。那么新建超源S超汇T，令degree[i]表示流入i的边的下界之和-从i流出的边的下界之和。
若degree[i]>0，则表示需要有额外degree[i]的流量流入i来达到流量平衡，那么从S向i连上界为degree[i]的边。
若degree[i]<0，则表示需要有额外degree[i]的流量从i流出来达到流量平衡，那么从i向T连上界为-degree[i]的边。
跑最大流就可以求出附加流。显然maxflow<=sigma(degree[i])。如果maxflow=sigma(degree[i])，那么有可行流。
#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,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()
{
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)
{
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));
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()
{
while (bfs())
{
memcpy(cur,p,sizeof(p));
ans+=work(S,inf);
}
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("loj115.in","r",stdin);
freopen("loj115.out","w",stdout);
const char LL[]="%I64d";
#else
const char LL[]="%lld";
#endif
n=read(),m=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]);
dinic();
if (ans<tot) cout<<"NO";
else
{
cout<<"YES\n";
for (int i=;i<=m;i++)
printf("%d\n",edge[i-<<].flow+l[i]);
}
return ;
}