CodeForces - 1000E ：We Need More Bosses（无向图缩点+树的直径）

Your friend is developing a computer game. He has already decided how the game world should look like — it should consist of two-way passages. The passages are designed in such a way that it should be possible to get from any location to any other location.

Of course, some passages should be guarded by the monsters (if you just can go everywhere without any difficulties, then it's not fun, right?). Some crucial passages will be guarded by really fearsome monsters, requiring the hero to prepare for battle and designing his own tactics of defeating them (commonly these kinds of monsters are called bosses). And your friend wants you to help him place these bosses.

The game will start in location t.

Input

The first line contains two integers n−1≤m≤3⋅105) — the number of locations and passages, respectively.

Then x≠y) describing the endpoints of one of the passages.

It is guaranteed that there is no pair of locations directly connected by two or more passages, and that any location is reachable from any other location.

Output

Print one integer — the maximum number of bosses your friend can place, considering all possible choices for t.

Examples

Input

Output

Input

Output

题意：一条路径上必经的边为关键边，现在让你找一条路径，使得其关键边最多，输出最多的数量。

思路：如果一条路径上面有环，那么这个环的任意一条边都不是关键边。所以我们缩点，那么就算在一棵树上找最多的关键边，显然就算求直径。

#include<bits/stdc++.h>
#define ll long long 
using namespace std;
const int maxn=600010;
int From[maxn],Laxt[maxn],To[maxn<<2],Next[maxn<<2],cnt;
int low[maxn],dfn[maxn],times,q[maxn],head,scc_cnt,scc[maxn];
vector<int>G[maxn];
int dis[maxn],S,T,ans;
void add(int u,int v)
{
    Next[++cnt]=Laxt[u]; From[cnt]=u;
    Laxt[u]=cnt; To[cnt]=v;
}
void tarjan(int u,int fa)
{
    dfn[u]=low[u]=++times; 
    q[++head]=u;
    for(int i=Laxt[u];i;i=Next[i]){
        if(To[i]==fa) continue;
        if(!dfn[To[i]]) {
              tarjan(To[i],u);
              low[u]=min(low[u],low[To[i]]);
        }
        else low[u]=min(low[u],dfn[To[i]]);
    }
    if(low[u]==dfn[u]){
        scc_cnt++;
        while(true){
            int x=q[head--];
            scc[x]=scc_cnt;
            if(x==u) break;
        }
    }
}
void dfs(int u,int f)
{
    dis[u]=dis[f]+1;
    for(int i=0;i<G[u].size();i++){
        if(G[u][i]!=f) dfs(G[u][i],u);
    }
}
int main()
{
    int N,M,u,v,i,j;
    scanf("%d%d",&N,&M);
    for(i=1;i<=M;i++){
        scanf("%d%d",&u,&v);
        add(u,v); add(v,u);
    }
    tarjan(1,0);
    for(i=1;i<=N;i++){
        for(j=Laxt[i];j;j=Next[j]){
           if(scc[i]!=scc[To[j]])
             G[scc[i]].push_back(scc[To[j]]);
        }
    }
    dfs(1,0);
    for(i=1;i<=scc_cnt;i++) if(dis[i]>dis[S]) S=i;
    dfs(S,0);
    for(i=1;i<=scc_cnt;i++) ans=max(ans,dis[i]-1);
    printf("%d\n",ans);
    return 0;
}

对于无向图的缩点：

void tarjan(int u,int fa)
{
    dfn[u]=low[u]=++times; 
    q[++head]=u;
    for(int i=Laxt[u];i;i=Next[i]){
        if(To[i]==fa) continue;
        if(!dfn[To[i]]) {
              tarjan(To[i],u);
              low[u]=min(low[u],low[To[i]]);
        }
        else low[u]=min(low[u],dfn[To[i]]);
    }
    if(low[u]==dfn[u]){
        scc_cnt++;
        while(true){
            int x=q[head--];
            scc[x]=scc_cnt;
            if(x==u) break;
        }
    }
}

对于有向图的缩点：二者唯一的区别就算有向图考虑横边，所以有个instack的判断。

void tarjan(int u)
{
    instk[u]=1;
    q[++head]=u;
    dfn[u]=low[u]=++times;
    for(int i=Laxt[u];i;i=Next[i]){
        int v=To[i];
        if(!dfn[v]) {
            tarjan(v);
            low[u]=min(low[u],low[v]);
        }
        else if(instk[v])low[u]=min(low[u],dfn[v]);//无向图与有向图的区别
    }
    if(dfn[u]==low[u]){
        scc_cnt++;
        while(true){
             int x=q[head--];
             scc[x]=scc_cnt;
             V[scc_cnt]+=w[x];
             instk[x]=0;
             if(x==u) break;
        }     
    }
}