时间复杂度:O(N)

void bfs(int u)
{
    memset( v , 0 , sizeof v );
    queue<int>q;
    dis[u] = 0;
    vis[u] = 1;
    q.push(u);
    while(q.size())
    {
        int x = q.front();
        q.pop();
        for(int i = head[x]; i ; i = next[i])
        {
            int y = ver[i];
            if(vis[y]) continue;
            dis[y] = dis[x] + edge[i];
            vis[y] = 1;
            q.push(y);
        }
    }

}

int main()
{ 
    bfs( 1 );
    for ( int i=1; i<=n; i++ )
        if ( s<dis[i] ) s=dis[i],rt1=i
    bfs( rt1 ); s=0; 
    for ( int i=1; i<=n; i++ )
        if ( s<dis[i] ) s=dis[i],rt2=i;
    //s为直径，rt1,rt2分别为直径两端
}