题意：给定上一个有容量和下界的网络，让你求出一组可行解。
析：先建立一个超级源点 s 和汇点 t ，然后在输入时记录到每个结点的下界的和，建边的时候就建立c - b的最后再建立 s 和 t ， 在建立时，如果 i 结点的输入的大于输出的，那么就是从 s 建立一条边，否则 i 与 t 建立，然后跑一次最大流，就OK了，注意求出的流量是没有下界，再加上下界的就好了。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(x,n) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 200 + 50;
const int mod = 1000;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
return r > 0 && r <= n && c > 0 && c <= m;
}
struct Edge{
int from, to, cap, flow;
};
struct Dinic{
int n, m, s, t;
vector<Edge> edges;
vector<int> G[maxn];
bool vis[maxn];
int d[maxn];
int cur[maxn];
void init(int n){
this-> n = n;
edges.clear();
for(int i = 0; i < n; ++i) G[i].clear();
}
void addEdge(int from, int to, LL cap){
edges.push_back((Edge){from, to, cap, 0});
edges.push_back((Edge){to, from, 0, 0});
m = edges.size();
G[from].push_back(m-2);
G[to].push_back(m-1);
}
bool bfs(){
memset(vis, 0, sizeof vis);
queue<int> q;
q.push(s);
d[s] = 0;
vis[s] = 1;
while(!q.empty()){
int x = q.front(); q.pop();
for(int i = 0; i < G[x].size(); ++i){
Edge &e = edges[G[x][i]];
if(!vis[e.to] && e.cap > e.flow){
vis[e.to] = 1;
d[e.to] = d[x] + 1;
q.push(e.to);
}
}
}
return vis[t];
}
int dfs(int x, int a){
if(x == t || a == 0) return a;
int flow = 0, f;
for(int &i = cur[x]; i < G[x].size(); ++i){
Edge &e = edges[G[x][i]];
if(d[x] + 1 == d[e.to] && (f = dfs(e.to, min(a, e.cap-e.flow))) > 0){
e.flow += f;
edges[G[x][i]^1].flow -= f;
flow += f;
a -= f;
if(a == 0) break;
}
}
return flow;
}
int maxFlow(int s, int t){
this->s = s; this->t = t;
int flow = 0;
while(bfs()){
memset(cur, 0, sizeof cur);
flow += dfs(s, INF);
}
return flow;
}
};
Dinic dinic;
int in[maxn*maxn], out[maxn*maxn];
int B[maxn*maxn];
int main(){
scanf("%d %d", &n, &m);
int s = 0, t = n + 1;
for(int i = 0; i < m; ++i){
int u, v, b, c;
scanf("%d %d %d %d", &u, &v, &b, &c);
dinic.addEdge(u, v, c - b);
B[i] = b;
in[v] += b; out[u] += b;
}
int ans = 0;
for(int i = 1; i <= n; ++i){
int c = in[i] - out[i];
if(c > 0) dinic.addEdge(s, i, c), ans += c;
else dinic.addEdge(i, t, -c);
}
if(dinic.maxFlow(s, t) != ans){ puts("NO"); return 0; }
puts("YES");
for(int i = 0; i < m; ++i)
printf("%d
", dinic.edges[i<<1].flow + B[i]);
return 0;
}