The terrorist group leaded by a well known international terrorist Ben Bladen is buliding a nuclear reactor to produce plutonium for the nuclear bomb they are planning to create. Being the wicked computer genius of this group, you are responsible for developing the cooling system for the reactor.
The cooling system of the reactor consists of the number of pipes that special cooling liquid flows by. Pipes are connected at special points, called nodes, each pipe has the starting node and the end point. The liquid must flow by the pipe from its start point to its end point and not in the opposite direction.
Let the nodes be numbered from 1 to N. The cooling system must be designed so that the liquid is circulating by the pipes and the amount of the liquid coming to each node (in the unit of time) is equal to the amount of liquid leaving the node. That is, if we designate the amount of liquid going by the pipe from i-th node to j-th as fij, (put fij= 0 if there is no pipe from node i to node j), for each i the following condition must hold:
fi,1+fi,2+…+fi,N = f1,i+f2,i+…+fN,i
Each pipe has some finite capacity, therefore for each i and j connected by the pipe must be fij <= cij where cij is the capacity of the pipe. To provide sufficient cooling, the amount of the liquid flowing by the pipe going from i-th to j-th nodes must be at least lij, thus it must be fij >= lij.
Given cij and lij for all pipes, find the amount fij, satisfying the conditions specified above.
This problem contains multiple test cases!
The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.
The output format consists of N output blocks. There is a blank line between output blocks.
Input
The first line of the input file contains the number N (1 <= N <= 200) – the number of nodes and and M – the number of pipes. The following M lines contain four integer number each – i, j, lij and cij each. There is at most one pipe connecting any two nodes and 0 <= lij <= cij <= 10^5 for all pipes. No pipe connects a node to itself. If there is a pipe from i-th node to j-th, there is no pipe from j-th node to i-th.
Output
On the first line of the output file print YES if there is the way to carry out reactor cooling and NO if there is none. In the first case M integers must follow, k-th number being the amount of liquid flowing by the k-th pipe. Pipes are numbered as they are given in the input file.
Sample Input
2
4 6
1 2 1 2
2 3 1 2
3 4 1 2
4 1 1 2
1 3 1 2
4 2 1 2
4 6
1 2 1 3
2 3 1 3
3 4 1 3
4 1 1 3
1 3 1 3
4 2 1 3
Sample Input
NO
YES
1
2
3
2
1
1
题意：
给n个点，及m根pipe，每根pipe用来流躺液体的，单向的，每时每刻每根pipe流进来的物质要等于流出去的物质，要使得m条pipe组成一个循环体，里面流躺物质。
并且满足每根pipe一定的流量限制，范围为[Li,Ri].即要满足每时刻流进来的不能超过Ri(最大流问题)，同时最小不能低于Li。(转自hzwer)
/*
无源汇上下界可行流.
建立源汇点.
计算出每个点进出流量的流量差s[i]=out[e]-in[e].
然后如果s[i]>0 则把流量s[i]导给T.
如果s[i]<0 则把从S流量补一条流量为-s[i]的边.
这样的弧我们称为必要弧.
然后想当与把下界分离开来.
若由S发出的弧(到达T的弧)都满流.
即这些弧的流量和等于最大流则为可行.
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#define MAXN 110
#define INF 1e9
using namespace std;
int n,m,S,T,cut=1,head[MAXN],s[MAXN],fa[MAXN],ans,sum,low[MAXN],dis[MAXN],b[MAXN];
struct data{int u,v,next,c;}e[MAXN*MAXN];
queue<int>q;
int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9') x=x*10+ch-48,ch=getchar();
return x*f;
}
void add(int u,int v,int c)
{
e[++cut].u=u;e[cut].v=v;e[cut].c=c;e[cut].next=head[u];head[u]=cut;
e[++cut].u=v;e[cut].v=u;e[cut].c=0;e[cut].next=head[v];head[v]=cut;
}
bool bfs()
{
q.push(S);
for(int i=0;i<=T;i++) dis[i]=-1,b[i]=0;dis[S]=0;
while(!q.empty())
{
int u=q.front();q.pop();b[u]=0;
for(int i=head[u];i;i=e[i].next)
{
int v=e[i].v;
if(dis[v]==-1&&e[i].c)
{
dis[v]=dis[u]+1;fa[v]=i;
if(!b[v]) b[v]=1,q.push(v);
}
}
}
return dis[T]!=-1;
}
int dfs(int u,int y)
{
if(u==T) return y;
int rest=0;
for(int i=head[u];i&&rest<y;i=e[i].next)
{
int v=e[i].v;
if(dis[v]==dis[u]+1&&e[i].c)
{
int x=dfs(v,min(y-rest,e[i].c));
e[i].c-=x;
e[i^1].c+=x;
rest+=x;
}
}
if(!rest) dis[u]=-1;
return rest;
}
int dinic()
{
ans=0;
while(bfs())
ans+=dfs(S,INF);
return ans;
}
int main()
{
int x,y,t,min1,max1;
t=read();
while(t--)
{
cut=1;sum=0;
memset(head,0,sizeof head);
memset(low,0,sizeof low);
memset(s,0,sizeof s);
n=read();m=read();S=n+1,T=n+2;
for(int i=1;i<=m;i++)
{
x=read(),y=read(),low[i]=read(),max1=read();
s[x]+=low[i],s[y]-=low[i];
add(x,y,max1-low[i]);
}
for(int i=1;i<=n;i++)
{
if(s[i]>0) add(i,T,s[i]),sum+=s[i];//导流.
else if(s[i]<0) add(S,i,-s[i]);//补流.
}
if(sum==dinic())
{
printf("YES\n");
for(int i=1;i<=m;i++)
printf("%d ",low[i]+e[(i<<1)^1].c);
}
else printf("NO\n");
}
return 0;
}