Codeforces 732F tarjan

Tourist Reform

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Berland is a tourist country! At least, it can become such — the government of Berland is confident about this.

There are n cities in Berland, some pairs of which are connected by two-ways roads. Each road connects two different cities. In Berland there are no roads which connect the same pair of cities. It is possible to get from any city to any other city using given two-ways roads.

According to the reform each road will become one-way. It will be oriented to one of two directions.

To maximize the tourist attraction of Berland, after the reform for each city i the value ri will be calculated. It will equal to the number of cities x for which there is an oriented path from the city i to the city x. In other words, ri will equal the number of cities which can be reached from the city i by roads.

The government is sure that tourist's attention will be focused on the minimum value of ri.

Help the government of Berland make the reform to maximize the minimum of ri.

Input

The first line contains two integers n, m (2 ≤ n ≤ 400 000, 1 ≤ m ≤ 400 000) — the number of cities and the number of roads.

The next m lines describe roads in Berland: the j-th of them contains two integers uj and vj (1 ≤ uj, vj ≤ n, uj ≠ vj), where uj and vj are the numbers of cities which are connected by the j-th road.

The cities are numbered from 1 to n. It is guaranteed that it is possible to get from any city to any other by following two-ways roads. In Berland there are no roads which connect the same pair of cities.

Output

In the first line print single integer — the maximum possible value min1 ≤ i ≤ n{ri} after the orientation of roads.

The next m lines must contain the description of roads after the orientation: the j-th of them must contain two integers uj, vj, it means that the j-th road will be directed from the city uj to the city vj. Print roads in the same order as they are given in the input data.

Example

input

7 9
4 3
2 6
7 1
4 1
7 3
3 5
7 4
6 5
2 5

output

4
4 3
6 2
7 1
1 4
3 7
5 3
7 4
5 6
2 5

题意：给你一个无向图，现在让你转化为有向图，也就是给每条边赋方向，使得每个点能到达的点数的最小值最大，求出这个值，然后输出路的方向。

题解：对于一个强联通分量，里面的点是可以相互到达的，对于两个相邻的强连通分量，设大小分别为x y，x>y,则如果x指向y，那么ans为y,否则答案为x，所以可得应该把小的分量指向大的分量，所以ans即为max(bcc i)，也就是强连通分量最大的那个值。此时其它的联通分量方向都指向这个最大的。

边的方向:对于一个强连通分量，随便找个点，然后dfs出去，每次都是to指向from，即可构成环，所以我们可以从最大的那个强联通分量dfs出去，这样就可以使得所有联通分量指向这个最大的。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int head[400005],cnt=0,n,m,bccnum,ef[800005],st[800005],index,top,dfn[800005],low[800005],bcc[800005],num[800005],now,ans,vis[800005];
struct node{
	int from,to,nex;
}edge[800005];
void add(int u,int v){
	edge[cnt].to=v;
	edge[cnt].from=u;
	edge[cnt].nex=head[u];
	head[u]=cnt++;
}
void tarjan(int root){//求强连通分量
    dfn[root]=low[root]=++index;
    st[++top]=root; 
    for(int i=head[root];~i;i=edge[i].nex){
        int v=edge[i].to;
        if(ef[i])continue;
        ef[i]=ef[i^1]=1;
        if(!dfn[v]){
            tarjan(v);
            low[root]=min(low[root],low[v]);
        }
        else low[root]=min(low[root],dfn[v]);
    }
    if(low[root]==dfn[root]){
        bccnum++;
        for(;;){
            int x=st[top--];
            bcc[x]=bccnum;
            num[bccnum]++;
            if(x==root)break;
        }
        if(num[bccnum]>ans){
        	now=root;//定义now为起始dfs点
			ans=max(ans,num[bccnum]);
        }
    }
}
void dfs(int root){
	if(vis[root])return ;//不加剪枝会t
	vis[root]=1;
	for(int i=head[root];~i;i=edge[i].nex){
		int v=edge[i].to;
		if(ef[i])continue;
		ef[i]=1;//代表走过
		ef[i^1]=2;//代表to指向from
		dfs(v);
	}
}
int main(){
	scanf("%d%d",&n,&m);
	int x,y,i,j;
	memset(head,-1,sizeof(head));
	for(i=1;i<=m;i++){
		scanf("%d%d",&x,&y);
		add(x,y);
		add(y,x);
	}
	tarjan(1);
	memset(ef,0,sizeof(ef));
	dfs(now);
	printf("%d\n",ans);	
	for(i=0;i<cnt;i++){
		if(ef[i]==2)printf("%d %d\n",edge[i].from,edge[i].to);
	}
	return 0;
}