time limit per test : 1 second
memory limit per test : 256 megabytes
分數:2000
You are given a directed graph with vertices and directed edges without self-loops or multiple edges.
Let’s denote the -coloring of a digraph as following: you color each edge in one of colors. The -coloring is good if and only if there no cycle formed by edges of same color.
Find a good -coloring of given digraph with minimum possible .
Input
The first line contains two integers and — the number of vertices and edges in the digraph, respectively.
Next lines contain description of edges — one per line. Each edge is a pair of integers and ≠ — there is directed edge from to in the graph.
It is guaranteed that each ordered pair appears in the list of edges at most once.
Output
In the first line print single integer — the number of used colors in a good -coloring of given graph.
In the second line print integers , where ci is a color of the -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 ).
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
題意:
給一個有向圖,給每條邊染色,使得每種顏色的邊不能構成環。求一種染色種數最少的方案。
題解:
最多隻會染上2種顏色
考慮dfs,每次出現返祖邊的時候返祖邊的顏色就是第二種顏色,否則的話全是第一種顏色。
那麼只要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;
}