脑残了，这题竟然都不会.
显然，把所有左右端点放在一起排序，然后取中位数是 $k=1$ 时最优的.
$k=2$ 的时候显然距离中点越近越好，所以将中点扔进去，然后枚举中间的分割点，分割点左右就变成两个子问题了.
动态求中位数的话用平衡树/权值线段树维护就行了.
code:
#include <bits/stdc++.h>
#define ll long long
#define lson s[x].ch[0]
#define rson s[x].ch[1]
#define N 200009
#define setIO(s) freopen(s".in","r",stdin)
using namespace std;
int tot,rt1,rt2,n,K,cnt;
ll seq[N<<1];
struct data
{
ll sum,w;
int ch[2],f,si;
}s[N<<2];
inline int get(int x) { return s[s[x].f].ch[1]==x; }
inline void pushup(int x)
{
s[x].si=s[lson].si+s[rson].si+1;
s[x].sum=s[lson].sum+s[rson].sum+s[x].w;
}
void rotate(int x)
{
int old=s[x].f,fold=s[old].f,which=get(x);
s[old].ch[which]=s[x].ch[which^1];
if(s[old].ch[which]) s[s[old].ch[which]].f=old;
s[x].ch[which^1]=old,s[old].f=x,s[x].f=fold;
if(fold) s[fold].ch[s[fold].ch[1]==old]=x;
pushup(old),pushup(x);
}
void splay(int x,int &tar)
{
int u=s[tar].f;
for(int fa;(fa=s[x].f)!=u;rotate(x))
if(s[fa].f!=u) rotate(get(fa)==get(x)?fa:x);
tar=x;
}
int getkth(int x,int kth)
{
while(1)
{
if(s[lson].si+1==kth) break;
else if(kth<=s[lson].si) x=lson;
else kth-=(s[lson].si+1),x=rson;
}
return x;
}
void ins(int &x,int fa,int v)
{
if(!x)
{
x=++tot,s[x].w=v,s[x].f=fa,pushup(x);
return;
}
ins(s[x].ch[v>s[x].w],x,v),pushup(x);
}
int find(int x,int v)
{
while(1)
{
if(s[x].w==v) break;
else x=s[x].ch[v>s[x].w];
}
return x;
}
void del(int v)
{
int x=find(rt2,v),l,r;
splay(x,rt2),l=s[x].ch[0],r=s[x].ch[1];
if(!l) s[r].f=0,rt2=r;
else if(!r) s[l].f=0,rt2=l;
else
{
while(s[l].ch[1]) l=s[l].ch[1];
splay(l,s[x].ch[0]);
s[l].f=0,s[l].ch[1]=r,s[r].f=l,rt2=l,pushup(rt2);
}
}
void build(int &x,int fa,int l,int r)
{
int mid=(l+r)>>1;
s[x=++tot].f=fa;
s[x].w=seq[mid];
if(mid>l) build(lson,x,l,mid-1);
if(r>mid) build(rson,x,mid+1,r);
pushup(x);
}
ll query(int &x)
{
int u=x;
int p=getkth(u,(s[u].si&1)?s[u].si/2+1:s[u].si/2);
splay(p,x);
ll ans=-s[lson].sum+(ll)s[x].w*s[lson].si-(ll)s[x].w*s[rson].si+s[rson].sum;
return ans;
}
struct node
{
ll l,r;
node(ll l=0,ll r=0):l(l),r(r){}
bool operator<(const node b) const { return (l+r)<(b.l+b.r); }
}nd[N];
int main()
{
// setIO("input");
scanf("%d%d",&K,&n);
char a[2],b[2];
ll ans=0,x,y,z;
for(int i=1;i<=n;++i)
{
scanf("%s%lld%s%lld",a,&x,b,&y);
if(x>y) swap(x,y);
if(a[0]==b[0]) ans+=abs(y-x);
else ++ans,seq[++cnt]=x,seq[++cnt]=y,nd[cnt/2]=node(x,y);
}
if(cnt==0) { printf("%lld\n",ans); return 0; }
sort(nd+1,nd+1+(cnt/2));
sort(seq+1,seq+1+cnt),build(rt2,0,1,cnt);
if(K==1) { printf("%lld\n",ans+query(rt2)); return 0; }
ll fin=100000000000000;
for(int i=1;i<cnt/2;++i)
{
del(nd[i].l);
del(nd[i].r);
ins(rt1,0,nd[i].l);
if(i%7==0) splay(tot,rt1);
ins(rt1,0,nd[i].r);
if(i%7==0) splay(tot,rt1);
fin=min(fin,query(rt1)+query(rt2));
}
printf("%lld\n",fin+ans);
return 0;
}
