Codeforces Round #550 (Div. 3) F. Graph Without Long Directed Paths（dfs）

F. Graph Without Long Directed Paths
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given a connected undirected graph consisting of n vertices and m edges. There are no self-loops or multiple edges in the given graph.

You have to direct its edges in such a way that the obtained directed graph does not contain any paths of length two or greater (where the length of path is denoted as the number of traversed edges).

Input
The first line contains two integer numbers n and m (2≤n≤2⋅105, n−1≤m≤2⋅105) — the number of vertices and edges, respectively.

The following m lines contain edges: edge i is given as a pair of vertices ui, vi (1≤ui,vi≤n, ui≠vi). There are no multiple edges in the given graph, i. e. for each pair (ui,vi) there are no other pairs (ui,vi) and (vi,ui) in the list of edges. It is also guaranteed that the given graph is connected (there is a path between any pair of vertex in the given graph).

Output
If it is impossible to direct edges of the given graph in such a way that the obtained directed graph does not contain paths of length at least two, print “NO” in the first line.

Otherwise print “YES” in the first line, and then print any suitable orientation of edges: a binary string (the string consisting only of ‘0’ and ‘1’) of length m. The i-th element of this string should be ‘0’ if the i-th edge of the graph should be directed from ui to vi, and ‘1’ otherwise. Edges are numbered in the order they are given in the input.

Example
inputCopy
6 5
1 5
2 1
1 4
3 1
6 1
outputCopy
YES
10100

题意

给你一个图，让你确定每一条边的方向，可以达到没有长度大于等于2的走法。
如果可以输出yes，并且输入每个边的方向
否则输出no

思路

这里可以想到的一点是，如果存在一个答案，那么把这个图的所有边都变成反向也是可以成立的。

可以想到的是如果一个点有边，那么所有的边的方向必然是朝向这个点或者都是从这个点发出，和这个点相邻的点和这个点就是不同的，比如第一个点的边的方向都是朝向这个点，那么和这个点相邻的点所有边的方向必然是发散的。所以这就转化成了一个染色问题。

code

#include<bits/stdc++.h>
#define pb push_back
using namespace std;
const int N=3e5+10;
int ar[N];
int x[N],y[N];
int vis[N],flag=0;
vector<int>G[N];
void dfs(int x,int pre){
    vis[x]=pre;
    for(int i=0;i<G[x].size();i++){
        int y=G[x][i];
        if(vis[y]==-1){
            dfs(y,1-pre);
        }
        else {
            if(vis[y]==vis[x]){
                flag=1;
                return ;
            }
        }
    }
}
int main()
{
    #ifdef local
        freopen("D://r.txt","r",stdin);
    #endif
    int n,m;
    cin>>n>>m;
    memset(vis,-1,sizeof vis);
    for(int i=0;i<m;i++){
        cin>>x[i]>>y[i];
        G[x[i]].pb(y[i]);
        G[y[i]].pb(x[i]);
    }
    dfs(1,1);
    if(flag==1){
        cout<<"NO"<<endl;
    }
    else {
        cout<<"YES"<<endl;
        for(int i=0;i<m;i++){
            if(vis[x[i]]){
                cout<<1;
            }
            else {
                cout<<0;
            }
        }
        cout<<endl;
    }
    return 0;
}