Educational Codeforces Round 72 (Rated for Div. 2)D. Coloring Edges（想法）

D. Coloring Edges

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a directed graph with nn vertices and mm directed edges without self-loops or multiple edges.

Let's denote the kk-coloring of a digraph as following: you color each edge in one of kk colors. The kk-coloring is good if and only if there no cycle formed by edges of same color.

Find a good kk-coloring of given digraph with minimum possible kk.

Input

The first line contains two integers nn and mm (2≤n≤50002≤n≤5000, 1≤m≤50001≤m≤5000) — the number of vertices and edges in the digraph, respectively.

Next mm lines contain description of edges — one per line. Each edge is a pair of integers uu and vv (1≤u,v≤n1≤u,v≤n, u≠vu≠v) — there is directed edge from uu to vv in the graph.

It is guaranteed that each ordered pair (u,v)(u,v) appears in the list of edges at most once.

Output

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

In the second line print mm integers c1,c2,…,cmc1,c2,…,cm (1≤ci≤k1≤ci≤k), where cici is a color of the ii-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 kk).

Examples

input

Copy

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

output

Copy

1
1 1 1 1 1 

input

Copy

3 3
1 2
2 3
3 1

output

Copy

2
1 1 2 

看到群里的解法

就是先用类似拓扑排序的思想

先判断一下是否存在环

如果存在环就是大☞小为1 反之为2

因为是一个环所有肯定会有两种情况

#include<bits/stdc++.h>
using namespace std;
vector<int>ve[5005];
int n,m;
int IN[5005];
int tuopu()
{
    queue<int>q;
    int ans=0;
    for(int i=1; i<=n; i++)
    {
        if(IN[i]==0)
        {
            q.push(i);
            ans++;
        }
    }
    while(q.size())
    {
        int u=q.front();
        q.pop();
        for(auto v:ve[u])
        {
            IN[v]--;
            if(IN[v]==0)
            {
                q.push(v);
                ans++;
            }
        }
    }
    return ans==n;
}
int tmp1[5005];
int tmp2[5005];
int main()
{
    scanf("%d%d",&n,&m);
    memset(IN,0,sizeof(IN));
    for(int i=1; i<=m; i++)
    {
        scanf("%d%d",&tmp1[i],&tmp2[i]);
        ve[tmp1[i]].push_back(tmp2[i]);
        IN[tmp2[i]]++;
    }
    if(tuopu())
    {
        printf("1
");
        for(int i=1; i<=m; i++)
        {
            printf("1%s",i==m?"
":" ");
        }
    }
    else
    {
        printf("2
");
        for(int i=1; i<=m; i++)
        {
            if(tmp1[i]>tmp2[i])
            {
                printf("1%s",i==m?"
":" ");
            }
            else
            {
                printf("2%s",i==m?"
":" ");
            }
        }
    }

    return 0;
}