P5149 [CF1707C] DFS Trees
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题目描述

You are given a connected undirected graph consisting of $n$ vertices and $m$ edges. The weight of the $i$-th edge is $i$.

Here is a wrong algorithm of finding a minimum spanning tree (MST) of a graph:

Each of the calls findMST(1), findMST(2), ..., findMST(n) gives you a spanning tree of the graph. Determine which of these trees are minimum spanning trees.

Input

The first line of the input contains two integers $n$, $m$ ($2\le n\le 10^5$, $n-1\le m\le 2\cdot 10^5$) — the number of vertices and the number of edges in the graph.

Each of the following $m$ lines contains two integers $u_i$ and $v_i$ ($1\le u_i, v_i\le n$, $u_i\ne v_i$), describing an undirected edge $(u_i,v_i)$ in the graph. The $i$-th edge in the input has weight $i$.

It is guaranteed that the graph is connected and there is at most one edge between any pair of vertices.

Output

You need to output a binary string $s$, where $s_i=1$ if findMST(i) creates an MST, and $s_i = 0$ otherwise.

输入格式

见题面 Input 部分

输出格式

见题面 Output 部分

样例输入
5 5 1 2 3 5 1 3 3 2 4 2
样例输出
01111
提示

Here is the graph given in the first example.

There is only one minimum spanning tree in this graph. A minimum spanning tree is $(1,2),(3,5),(1,3),(2,4)$ which has weight $1+2+3+5=11$.

Here is a part of the process of calling findMST(1):

  • reset the array vis and the edge set s;
  • calling dfs(1);
  • vis[1] := true;
  • iterate through each edge $(1,2),(1,3)$;
  • add edge $(1,2)$ into the edge set s, calling dfs(2): vis[2] := true iterate through each edge $(2,1),(2,3),(2,4)$; because vis[1] = true, ignore the edge $(2,1)$; add edge $(2,3)$ into the edge set s, calling dfs(3): ...

In the end, it will select edges $(1,2),(2,3),(3,5),(2,4)$ with total weight $1+4+2+5=12>11$, so findMST(1) does not find a minimum spanning tree.

It can be shown that the other trees are all MSTs, so the answer is 01111.