SGU 194带上下界的可行流。

194. Reactor Cooling

time limit per test: 0.5 sec.
memory limit per test: 65536 KB

input: standard
output: standard

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 f _ij, (put f _ij = 0 if there is no pipe from node i to node j), for each i the following condition must hold:


sum(j=1..N, f _ij) = sum(j=1..N, f _ji)


Each pipe has some finite capacity, therefore for each i and j connected by the pipe must be f _ij ≤ c _ij where c _ij 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 l _ij, thus it must be f _ij ≥ l _ij.

Given c _ij and l _ij for all pipes, find the amount f _ij, satisfying the conditions specified above.

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, l _ij and c _ij each. There is at most one pipe connecting any two nodes and 0 ≤ l _ij ≤ c _ij ≤ 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 test(s)

Input

Test #1

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

Test #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

Output

Test #1

NO

Test #2

YES
1
2
3
2
1

1

流网络里的边流量既有上界，又有下界。输出一组可行流。

代码：

/* ***********************************************
Author :rabbit
Created Time :2014/4/6 2:10:40
File Name :10.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
/*
对于有流量上下限的无源的网络流的可行流转化为一般的有源汇点的最大流来做 
(1)添加超级源点S和超级汇点T 
(2)对于原有的边(u,v,l(u,v),c(u,v))(l为流量下限,c为流量上限),添加边(u,v,0,c-l); 
(3)对于每个结点i,记w[i]=sum(l(u,i))-sum(l(i,v)); 
   若w[i]>0,添加边(S,i,w[i]),若w[i]<0,添加边(i,T,-w[i]); 
(4)求解S-T的最大流; 
(5)当且仅当S的出边和T的入边满流,原流量限制的无源网络流可行流有解; 
(6)一组可行流的解为: 
   对于每条流量边(u,v),可行流量为l(u,v)+其构造的图中的流量. 
*/ 

const int maxn=20100;
const int maxm=1002000;
struct Edge{
	int next,to,cap;
	Edge(){};
	Edge(int _next,int _to,int _cap){
		next=_next;to=_to;cap=_cap;
	}
}edge[maxm];
int head[maxn],tol,dep[maxn],gap[maxn];
void addedge(int u,int v,int flow){
    edge[tol]=Edge(head[u],v,flow);head[u]=tol++;
    edge[tol]=Edge(head[v],u,0);head[v]=tol++;
}
void bfs(int start,int end){
    memset(dep,-1,sizeof(dep));
    memset(gap,0,sizeof(gap));
    gap[0]++;int front=0,rear=0,Q[maxn];
    dep[end]=0;Q[rear++]=end;
    while(front!=rear){
        int u=Q[front++];
        for(int i=head[u];i!=-1;i=edge[i].next){
            int v=edge[i].to;if(dep[v]==-1)
                Q[rear++]=v,dep[v]=dep[u]+1,gap[dep[v]]++;
        }
    }
}
int sap(int s,int t,int N){
	int res=0;bfs(s,t);
	int cur[maxn],S[maxn],top=0,u=s,i;
	memcpy(cur,head,sizeof(head));
	while(dep[s]<N){
		if(u==t){
			int temp=INF,id;
		    for( i=0;i<top;i++)
			   if(temp>edge[S[i]].cap)
				   temp=edge[S[i]].cap,id=i;
		    for( i=0;i<top;i++)
			      edge[S[i]].cap-=temp,edge[S[i]^1].cap+=temp;
		    res+=temp;top=id;u=edge[S[top]^1].to;
		}
		if(u!=t&&gap[dep[u]-1]==0)break;
		for( i=cur[u];i!=-1;i=edge[i].next)
			if(edge[i].cap&&dep[u]==dep[edge[i].to]+1)break;
		if(i!=-1)cur[u]=i,S[top++]=i,u=edge[i].to;
		else{
			int MIN=N;
			for( i=head[u];i!=-1;i=edge[i].next)
				if(edge[i].cap&&MIN>dep[edge[i].to])
					MIN=dep[edge[i].to],cur[u]=i;
			--gap[dep[u]];++gap[dep[u]=MIN+1];
			if(u!=s)u=edge[S[--top]^1].to;
		}
	}
	return res;
}
int w[maxn],l[maxn];
int main()
{
     //freopen("data.in","r",stdin);
     //freopen("data.out","w",stdout);
     int n,m;
	 while(cin>>n>>m){
		 memset(head,-1,sizeof(head));tol=0;
		 memset(w,0,sizeof(w));
		 for(int i=0;i<m;i++){
			 int u,v,c;
			 scanf("%d%d%d%d",&u,&v,&l[i],&c);
			 addedge(u,v,c-l[i]);
			 w[u]-=l[i];
			 w[v]+=l[i];
		 }
		 int sum=0;
		 for(int i=1;i<=n;i++){
			 if(w[i]>0)addedge(0,i,w[i]),sum+=w[i];
			 if(w[i]<0)addedge(i,n+1,-w[i]);
		 }
		 int ans=sap(0,n+1,2*n+10);
		 if(ans!=sum)puts("NO");
		 else{
			 puts("YES");
			 for(int i=0;i<m;i++)printf("%d\n",edge[2*i+1].cap+l[i]);
		 }
	 }
     return 0;
}