题解：
由于线段可以出现平行于$y$轴的情况，
我们要拆点。
然后分情况讨论。
具体方法不赘述。
代码：
#include<cmath>
#include<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 2050
#define ll long long
const int inf = 0x3f3f3f3f;
const ll Inf = 0x3f3f3f3f3f3f3f3fll;
inline int rd()
{
int f=1,c=0;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){c=10*c+ch-'0';ch=getchar();}
return f*c;
}
int n,m,hed[N],cnt=-1,S,T,tot=0;
struct seg
{
int l,r;
ll w;
}s[N];
struct Pair
{
int x,id,k;
Pair(){}
Pair(int x,int i,int k):x(x),id(i),k(k){}
}p[N];
bool cmp(Pair a,Pair b)
{
if(a.x!=b.x)return a.x<b.x;
return a.k<b.k;
}
struct EG
{
int to,nxt;
ll w,c;
}e[N*N*4];
void ae(int f,int t,ll w,ll c)
{
e[++cnt].to = t;
e[cnt].nxt = hed[f];
e[cnt].w = w;
e[cnt].c = c;
hed[f] = cnt;
}
void AE(int f,int t,ll w,ll c)
{
ae(f,t,w,c);
ae(t,f,0,-c);
}
ll dep[N],fl[N];
int pre[N],fa[N];
bool vis[N];
queue<int>q;
bool spfa()
{
memset(dep,0x3f,sizeof(dep));
dep[S] = 0,fl[S] = inf,vis[S] = 1;
q.push(S);
while(!q.empty())
{
int u = q.front();
q.pop();
for(int j=hed[u];~j;j=e[j].nxt)
{
int to = e[j].to;
if(e[j].w&&dep[to]>dep[u]+e[j].c)
{
dep[to] = dep[u]+e[j].c;
fl[to] = min(fl[u],e[j].w);
pre[to] = j,fa[to] = u;
if(!vis[to])
{
vis[to] = 1;
q.push(to);
}
}
}
vis[u] = 0;
}
return dep[T]!=Inf;
}
ll mcmf()
{
ll ret = 0;
while(spfa())
{
ret+=fl[T]*dep[T];
int u = T;
while(u!=S)
{
e[pre[u]].w-=fl[T];
e[pre[u]^1].w+=fl[T];
u = fa[u];
}
}
return ret;
}
vector<ll>ve[N];
int main()
{
n = rd(),m = rd();
S = 0,T = 1;
memset(hed,-1,sizeof(hed));
int k = 0;
for(int x,y,i=1;i<=n;i++)
{
s[i].l=rd(),x=rd(),s[i].r=rd(),y=rd(),s[i].w=(ll)sqrt(1ll*(s[i].r-s[i].l)*(s[i].r-s[i].l)+1ll*(x-y)*(x-y));
if(s[i].l>s[i].r)
{
swap(s[i].l,s[i].r);
}
p[++tot] = Pair(s[i].l,i,0);
p[++tot] = Pair(s[i].r,i,1);
}
sort(p+1,p+1+tot,cmp);
for(int las=-1,i=1;i<=tot;i++)
{
if(las!=p[i].x)
{
las = p[i].x;
k++;
}
if(!p[i].k)s[p[i].id].l = k;
else
{
if(s[p[i].id].l!=k)AE(s[p[i].id].l<<1|1,k<<1,1,-s[p[i].id].w);
else AE(k<<1,k<<1|1,1,-s[p[i].id].w);
}
}
AE(S,2,m,0);AE(k<<1|1,T,m,0);
for(int i=1;i<k;i++)
AE(i<<1|1,(i+1)<<1,m,0);
for(int i=1;i<=k;i++)
AE(i<<1,i<<1|1,m,0);
printf("%lld\n",-mcmf());
return 0;
}