最长k可重线段集问题

题目描述

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;
}