Title:

Give n points and M pipes. Each pipe is used to flow liquid. One way, the material flowing in each pipe at every moment is equal to the material flowing out. Make m pipes form a cycle body, in which the material flows.

The flow limit of each pipe is [Li,Ri]. That is to say, the flow in at every time should not exceed RI (maximum flow problem), and the minimum flow should not be lower than Li.

Explanation:

The upper bound is ci, and the lower bound is bi.

The lower bound must be full, so for each edge, after removing the lower bound, its free flow is ci – bi.

Main idea: flow in at every point = flow out

For each point i, let

Mi= sum(i-point all the lower bound flows in) - sum(i-point all the lower bound flows out)

If Mi is greater than 0, it means that this point must also flow out the free flow of Mi, then we connect the edge of a Mi from the source point to this point.

If Mi is less than 0, it means that this point must also flow in the free flow of Mi, then we can connect the edge of Mi to the confluence point from this point.

If the maximum flow of S - > t is found, check whether the flow is full (the adjacent edges of s are full).

Full flow has solution, otherwise there is no solution.

CODE:

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #define inf 0x7fffffff
 5 #define T 201
 6 using namespace std;
 7 inline int read()
 8 {
 9     int x=0,f=1;char ch=getchar();
10     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
11     while(ch>='0'&&ch<='9'){x*=10;x+=ch-'0';ch=getchar();}
12     return x*f;
13 }
14 int n,m,cnt;
15 int head[205],cur[205],h[205],q[205],in[205];
16 int low[100005];
17 struct data{int to,next,v;}e[100005];
18 void ins(int u,int v,int w)
19 {e[++cnt].to=v;e[cnt].next=head[u];head[u]=cnt;e[cnt].v=w;}
20 void insert(int u,int v,int w)
21 {ins(u,v,w);ins(v,u,0);}
22 bool bfs()
23 {
24      for(int i=1;i<=T;i++)h[i]=-1;
25      int t=0,w=1;q[0]=0;h[0]=0;
26      while(t!=w)
27      {
28           int now=q[t];t++;
29           for(int i=head[now];i;i=e[i].next)
30               if(e[i].v&&h[e[i].to]==-1)
31               {
32                    h[e[i].to]=h[now]+1;
33                    q[w++]=e[i].to;
34               }
35      }
36      if(h[T]==-1)return 0;
37      return 1;
38 }
39 int dfs(int x,int f)
40 {
41      if(x==T)return f;
42      int w,used=0;
43      for(int i=cur[x];i;i=e[i].next)
44           if(h[e[i].to]==h[x]+1)
45           {
46                w=f-used;w=dfs(e[i].to,min(e[i].v,w));
47                e[i].v-=w;if(e[i].v)cur[x]=i;e[i^1].v+=w;
48                used+=w;if(used==f)return f;
49           }
50      if(!used)h[x]=1;
51      return used;
52 }
53 void dinic()
54 {while(bfs()){for(int i=0;i<=T;i++)cur[i]=head[i];dfs(0,inf);}}
55 void build()
56 {
57      for(int i=1;i<=n;i++)
58          if(in[i]>0)insert(0,i,in[i]);
59          else insert(i,T,-in[i]);
60 }
61 bool jud()
62 {
63      for(int i=head[0];i;i=e[i].next)
64          if(e[i].v)return 0;
65      return 1;
66 }
67 int main()
68 {
69      int test=read();
70      while(test--)
71      {
72           cnt=1;
73           memset(head,0,sizeof(head));
74           memset(in,0,sizeof(in));
75           n=read();m=read();
76           for(int i=1;i<=m;i++)
77           {
78               int u,v,w;
79               u=read();v=read();low[i]=read();w=read();
80               in[u]-=low[i];in[v]+=low[i];
81               insert(u,v,w-low[i]);
82           }
83           build();
84           dinic();
85           if(!jud())printf("NO\n");
86           else 
87           {
88                printf("YES\n");
89                for(int i=1;i<=m;i++)
90                    printf("%d\n",e[(i<<1)^1].v+low[i]);
91           }
92           puts("");
93      }
94      return 0;
95 }