求一个无源无汇的有上下界的最大流。

把每条边的边容量重设为上界-下界，记录每个点的最小进入in和最小输出流out，重设一个s、t，将s、t连向每一个点，边容量分别为每个点的in和out，从s到t跑最大流即

可。如果s的每条边满流，则有解，否则无解。

#include <cstdio>
#include <algorithm>
#include <queue>
using namespace std;
const int MAXN = 209,INF = 1e9;
int n,m;
struct Edge
{
	int v,ne;
	int f,c,l;		
}edge[MAXN*MAXN];
int edgehead[MAXN],p = 1;
int sum,in[MAXN],out[MAXN];
int dis[MAXN];

void add(int u,int v,int f,int l)
{
	edge[p].v = v;edge[p].c = edge[p].f = f;edge[p].l = l;
	edge[p].ne = edgehead[u];edgehead[u] = p++;	
}

int adv(int k) {return ((k-1)^1)+1;}

bool bfs(int s,int t)
{
	for(int i = 1; i <= n; ++i) dis[i] = INF;
	dis[t] = 0;
	queue<int> q;
	q.push(t);
	while(!q.empty())
	{
		int u = q.front();q.pop();
		for(int i = edgehead[u]; i; i = edge[i].ne)
		{
			int v = edge[i].v,f = edge[adv(i)].f;
 			if(f && dis[v] > dis[u]+1)
 			{
 				dis[v] = dis[u]+1;
				q.push(v);	
 			}
		}
	}
	return dis[s] != INF;
}

int dfs(int now,int t,int flow)
{
	if(now == t || flow == 0) return flow;
	int re = 0,tmp;
	for(int i = edgehead[now]; i; i = edge[i].ne)
	{
		int to = edge[i].v;
		if(dis[to] == dis[now]-1 && edge[i].f && (tmp = dfs(to,t,min(flow,edge[i].f))))
		{
			re += tmp;flow -= tmp;
			edge[i].f -= tmp;edge[adv(i)].f += tmp;
			if(flow == 0) break;
		}
	}
	dis[now] = INF;
	return re;
}

int dinic(int s,int maxf)
{
	int re = 0;
	while(bfs(s,t))
		re += dfs(s,maxf);
	return re;
}

int main()
{
	scanf("%d%d",&n,&m);
	for(int i = 1; i <= m; ++i)
	{
		int a,b,d;
		scanf("%d%d%d%d",&a,&b,&c,&d);
		in[b] += c;out[a] += c;sum += c;
		add(a,d-c,c);
		add(b,a,0);	
	}
	for(int i = 1; i <= n; ++i)
	{
		if(in[i]) add(n+1,i,in[i],0),add(i,n+1,0);
		if(out[i]) add(i,n+2,out[i],add(n+2,0);	
	}
	n += 2;
	if(dinic(n-1,n,INF) != sum)
	{
		puts("NO");
		goto end;
	}
	puts("YES");
	for(int i = 1; i <= m; ++i)
		printf("%d\n",edge[2*i-1].c-edge[2*i-1].f+edge[2*i-1].l);
	
	end:;
	return 0;	
}