F. Graph Without Long Directed Paths
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given a connected undirected graph consisting of nn vertices and mm 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 nn and mm (2≤n≤2⋅1052≤n≤2⋅105, n−1≤m≤2⋅105n−1≤m≤2⋅105) — the number of vertices and edges, respectively.
The following mm lines contain edges: edge ii is given as a pair of vertices uiui, vivi (1≤ui,vi≤n1≤ui,vi≤n, ui≠viui≠vi). There are no multiple edges in the given graph, i. e. for each pair (ui,viui,vi) there are no other pairs (ui,viui,vi) and (vi,uivi,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 mm. The ii-th element of this string should be '0' if the ii-th edge of the graph should be directed from uiui to vivi, and '1' otherwise. Edges are numbered in the order they are given in the input.
Example
input
Copy
6 5 1 5 2 1 1 4 3 1 6 1
output
Copy
YES 10100
Note
The picture corresponding to the first example:
此题并不难
将点分成两类
入点 和出点 入点所有的边都是入度 出点相反
由题意可知 两个相邻点必为两种类型不同的点
因此 此题只需要跑一遍dfs看是否有冲突的现象即可
ac代码
#include <cstdio>
#include <cstring>
#include <cmath>
#include <ctime>
#include <cstdlib>
#include <queue>
#include <stack>
#include <string>
#include <iostream>
#include <algorithm>
#include <map>
#include<algorithm>
#include <set>
using namespace std;
int n,m;
int a[200005];
int first[400005];
int to[400005];
int nxt[400005];
int stu[200005];
int flag=0;
void dfs(int x,int lst)
{
if(stu[x]==-1) stu[x]=lst^1;
else if(stu[x]==lst)
{
flag=-1;
printf("NO
");
exit(0);
}
else
{
return ;
}
for(int i=first[x]; i!=-1; i=nxt[i])
{
dfs(to[i],stu[x]);
}
return ;
}
int main()
{
memset(stu,-1,sizeof(stu));
memset(first,-1,sizeof(first));
memset(to,-1,sizeof(to));
memset(nxt,-1,sizeof(nxt));
scanf("%d%d",&n,&m);
int cnt=0;
for(int i=1; i<=m; i++)
{
int temp1;
int temp2;
scanf("%d%d",&temp1,&temp2);
a[i]=temp1;
cnt++;
nxt[cnt]=first[temp1];
to[cnt]=temp2;
first[temp1]=cnt;
cnt++;
nxt[cnt]=first[temp2];
to[cnt]=temp1;
first[temp2]=cnt;
}
dfs(1,0);
printf("YES
");
for(int i=1;i<=m;i++)
{
printf("%d",stu[a[i]]);
}
return 0;
}