題目：

n個點m條邊，無窮多個公司，給每個公司最多分配兩條邊，每條邊都要被分配出去，且每個點相連的所有的邊所屬的公司數不超過k。輸出一種邊分配方案。

分析：

記錄每個點的度數，如果一個點的度數d[i] <= k，那麼這個點相連的邊無需合併，可直接給每個邊分配一個公司（因為公司有無窮多個）
如果一個點的度數d[i] > k，那麼這個點的邊至少要合併 2*(d[i]-k)條，而每條邊只能被合併一次。這樣便可最大流判定是否有解，源點S向每個城市點建流量為max(0, 2*(d[i]-k))的邊，城市點向對應的道路點連容量為1的邊，每個道路點向匯點T建立容量為1的邊。

程式碼：

#include <bits/stdc++.h>
using namespace std;
#define ms(a,b) memset(a,b,sizeof(a))
#define lson rt*2,l,(l+r)/2
#define rson rt*2+1,(l+r)/2+1,r
typedef unsigned long long ull;
typedef long long ll;
const int MAXN = 1205;
const int MAXM = 5005;
const int INF = 0x3f3f3f3f;
struct Edge {
	int to, next, cap, flow;
} edge[MAXM];
int tot, head[MAXN];
int Q[MAXN], cur[MAXN], dep[MAXN];
int n, m, S, T, N, to[MAXN];
void init() {
	tot = 2;
	memset(head, -1, sizeof head);
}
void addedge(int u, int v, int w, int rw = 0) {
	edge[tot].to = v; edge[tot].cap = w; edge[tot].flow = 0;
	edge[tot].next = head[u]; head[u] = tot++;
	edge[tot].to = u; edge[tot].cap = rw; edge[tot].flow = 0;
	edge[tot].next = head[v]; head[v] = tot++;
}
bool bfs(int s, int t, int n) {
	int Front = 0, tail = 0;
	memset(dep, -1, sizeof(dep[0]) * (n + 1));
	dep[s] = 0;
	Q[tail++] = s;
	while (Front < tail) {
		int u = Q[Front++];
		for (int i = head[u]; i != -1; i = edge[i].next) {
			int v = edge[i].to;
			if (edge[i].cap > edge[i].flow && dep[v] == -1) {
				dep[v] = dep[u] + 1;
				if (v == t) return true;
				Q[tail++] = v;
			}
		}
	}
	return false;
}
int dfs(int u, int f) {
	if (u == T)	return f;
	int used = 0, rflow = 0;
	for (int i = cur[u]; i != -1; i = edge[i].next) {
		cur[u] = i;
		int v = edge[i].to, w = edge[i].cap - edge[i].flow;
		if (w > 0 && dep[v] == dep[u] + 1) {
			if ((rflow = dfs(v, min(w, f - used)))) {
				used += rflow;
				edge[i].flow += rflow;
				edge[i ^ 1].flow -= rflow;
				if (used == f)	break;
			}
		}
	}
	if (!used)	dep[u] = -1;
	return used;
}
int dinic(int s, int t, int n) {
	int maxflow = 0;
	while (bfs(s, t, n)) {
		for (int i = 0; i <= n; i++)	cur[i] = head[i];
		maxflow += dfs(s, INF);
	}
	return maxflow;
}
int col[MAXN], d[MAXN], k;
int main() {
	freopen("1.txt","r",stdin);
	ios::sync_with_stdio(false);
	int kase;
	cin >> kase;
	while (kase--) {
		cin >> n >> m >> k;
		init();
		S = 0, T = n + m + 1, N = n + m + 1;
		for (int i = 0; i <= max(n,m); i++) d[i] = 0, col[i] = 0;
		for (int i = 1; i <= m; i++) {
			int a, b; cin >> a >> b;
			d[a]++, d[b]++;
			addedge(a, n + i, 1); addedge(b, n + i, 1);
			addedge(n + i, T, 1);
		}
		int sum = 0;
		for (int i = 1; i <= n; i++) {
			if (d[i] > k) {
				addedge(S, i, 2 * (d[i] - k));
				sum += 2 * (d[i] - k); 
			}
		}
		if (sum == dinic(S, T, N)) {
			int now = 0;
			vector<int> tp;
			for(int i=1;i<=n;i++) {
				for(int k=head[i];k!=-1;k=edge[k].next) {
					if(edge[k].flow == 1)	tp.push_back(edge[k].to);
				}
			}
			for(int i=0;i<(int)tp.size();i+=2){
				col[tp[i]-n] = col[tp[i+1]-n] = ++now;
			}
			for(int i=1;i<=m;i++) if(col[i] == 0) col[i] = ++now;
		} 
		for (int i = 1; i <= m; i++) cout << col[i] << " \n"[i == m];
	}
	return 0;
}