Reactor Cooling

Time Limit: 5 Seconds      Memory Limit: 32768 KB      Special Judge

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

題意：給以個流量圖，每個邊的流量必須為[l,r]，沒有源匯點，問是否能跑出一個可行流並輸出每條邊的流量。

無源匯點有下界可行流建圖方法：

對於每個點，求出流入流量下界之和和流出流量下界之和，所有邊u->v的流量下界變位0，上界變位high-low。但是這樣不能保證流量守恆，所以需要建立附加流量，設源點s和匯點t，對於所有下界流入量大於流出量的點，s與之建邊，流量為兩者之車，對於所有下界流入量小於流出量的點，與t建邊，流量為兩者之差。此時跑一邊最大流，如果從源點流出的流量之和等於最大流，就說明可以流量守恆即存在可行流，那麼沒條邊的流量等於該條邊的流量下界+附加流跑出的流量。

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<string>
#include<map>
#include<queue>
using namespace std;
const int maxn = 205;
const int maxm = 50005;
const int INF = 1e9 + 7;
struct node
{
	int u, v, flow, next, low;
}edge[maxm];
int n, m, s, t, cnt;
int head[maxn], dis[maxn], pre[maxm], cur[maxm], in[maxn], out[maxn];
char str[205][205];
void init()
{
	cnt = s = 0, t = n + 1;
	memset(head, -1, sizeof(head));
	memset(in, 0, sizeof(in));
	memset(out, 0, sizeof(out));
}
void add(int u, int v, int flow, int low)
{
	edge[cnt].u = u, edge[cnt].v = v, edge[cnt].flow = flow, edge[cnt].low = low;
	edge[cnt].next = head[u], head[u] = cnt++;
	edge[cnt].u = v, edge[cnt].v = u, edge[cnt].flow = 0, edge[cnt].low = low;
	edge[cnt].next = head[v], head[v] = cnt++;
}
int bfs()
{
	queue<int>q;
	memset(dis, -1, sizeof(dis));
	dis[s] = 0;
	q.push(s);
	while (!q.empty())
	{
		int u = q.front();q.pop();
		for (int i = head[u];i != -1;i = edge[i].next)
		{
			int v = edge[i].v;
			if (dis[v] == -1 && edge[i].flow)
			{
				dis[v] = dis[u] + 1;
				q.push(v);
			}
		}
	}
	if (dis[t] == -1) return 0;
	return 1;
}
int dfs(int u, int flow)
{
	if (u == t) return flow;
	for (int &i = cur[u];i != -1;i = edge[i].next)
	{
		int v = edge[i].v;
		if (dis[v] == dis[u] + 1 && edge[i].flow)
		{
			int d = dfs(v, min(edge[i].flow, flow));
			if (d > 0)
			{
				edge[i].flow -= d;
				edge[i ^ 1].flow += d;
				return d;
			}
		}
	}
	return 0;
}
int dinic()
{
	int ans = 0, d;
	while (bfs())
	{
		for (int i = s;i <= t;i++) cur[i] = head[i];
		while (d = dfs(s, INF))
			ans += d;
	}
	return ans;
}
int main()
{
	int i, j, k, sum, x, T;
	int u, v, low, high;
	scanf("%d", &T);
	while (T--)
	{
		scanf("%d%d", &n, &m);
		init();sum = 0;
		for (i = 1;i <= m;i++)
		{
			scanf("%d%d%d%d", &u, &v, &low, &high);
			add(u, v, high - low, low);
			in[v] += low, out[u] += low;
		}
		for (i = 1;i <= n;i++)
		{
			if (in[i] > out[i])
			{
				sum += in[i] - out[i];
				add(s, i, in[i] - out[i], 0);
			}
			else add(i, t, out[i] - in[i], 0);
		}
		if (dinic() != sum)
			printf("NO\n");
		else
		{
			printf("YES\n");
			for (i = 0;i < m * 2;i += 2)
				printf("%d\n", edge[i ^ 1].flow + edge[i].low);
		}
	}
	return 0;
}