time limit per test : 1 second
memory limit per test : 256 megabytes

分数：2000

You are given a directed graph with $n$ vertices and $m$ directed edges without self-loops or multiple edges.
Let’s denote the $k$-coloring of a digraph as following: you color each edge in one of $k$ colors. The $k$-coloring is good if and only if there no cycle formed by edges of same color.
Find a good $k$-coloring of given digraph with minimum possible $k$.

Input

The first line contains two integers $n$ and $m (2≤n≤5000, 1≤m≤5000)$ — the number of vertices and edges in the digraph, respectively.
Next $m$ lines contain description of edges — one per line. Each edge is a pair of integers $u$ and $v(1≤u,v≤n, u$≠ $v)$ — there is directed edge from $u$ to $v$ in the graph.
It is guaranteed that each ordered pair $(u,v)$ appears in the list of edges at most once.

Output

In the first line print single integer $k$— the number of used colors in a good $k$-coloring of given graph.

In the second line print $m$ integers $c_1,c_2,…,c_m (1≤c_i≤k)$, where ci is a color of the $i$-th edge (in order as they are given in the input).
If there are multiple answers print any of them (you still have to minimize $k$).

Examples
Input

4 5
1 2
1 3
3 4
2 4
1 4

Outpu

1
1 1 1 1 1 

Input

3 3
1 2
2 3
3 1

Output

2
1 1 2 

题意：
给一个有向图，给每条边染色，使得每种颜色的边不能构成环。求一种染色种数最少的方案。

题解：
最多只会染上2种颜色
考虑dfs，每次出现返祖边的时候返祖边的颜色就是第二种颜色，否则的话全是第一种颜色。
那么只要dfs即可。

#include<bits/stdc++.h>
#define ll long long
#define pa pair<int,int>
using namespace std;
int n,m;
vector<pa>G[5004];
int ans[5004];
bool flag,vis[5004],inq[5004];
void dfs(int x){
    inq[x]=1;
    vis[x]=1;
    for(pa now:G[x]){
        int to=now.first,s=now.second;
        ans[s]=1;
        if(!vis[to]){
            dfs(to);
        }
        else if(inq[to]){
            ans[s]=2;
            flag=1;
        }
        else{
            ans[s]=1;
        }
    }
    inq[x]=0;
}
int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++){
        int u,v;scanf("%d%d",&u,&v);
        G[u].push_back({v,i});
    }
    flag=0;
    for(int i=1;i<=n;i++){
        if(!vis[i])dfs(i);
    }
    if(!flag){
        printf("%d\n",1);
        for(int i=1;i<=m;i++){
            printf("%d ",ans[i]);
        }
        puts("");
    }
    else{
        printf("%d\n",2);
        for(int i=1;i<=m;i++){
            printf("%d ",ans[i]);
        }
        puts("");
    }
    return 0;
}