题目链接
题意
x0y平面上n条开线段，求一个线段子集，使得它的线段长度和最长，且对于每条垂直于x轴的线段都不超过k条线段与之相交。
思路
类似最长k可重区间集问题，因为只对x轴上的点有限制，一样的，离散化后建边，跑最大费用最大流，需要注意的是这里是开线段，有个神奇的处理就是把l改为l * 2+1，r改为r * 2。先对x离散化，处理出费用，也就是线段的长度，对于每个相邻的点连上容量为INF，费用为0的边，对于每一条线段，左端点与右端点连上容量为1,，费用为线段长度的边，源点连第一个节点，最后一个节点连汇点。然后跑最大费用最大流。

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
const int N = 5e3 + 7;
const int M = 1e4 + 7;
const int maxn = 5e3 + 7;
typedef long long ll;
int maxflow, mincost;
struct Edge {
    int from, to, cap, flow, cost;
    Edge(int u, int v, int ca, int f, int co):from(u), to(v), cap(ca), flow(f), cost(co){};
};
struct MCMF
{
    int n, m, s, t;
    vector<Edge> edges;
    vector<int> G[N];
    int inq[N], d[N], p[N], a[N];//是否在队列 距离 上一条弧 可改进量
    void init(int n) {
        this->n = n;
        for (int i = 0; i < n; i++) G[i].clear();
        edges.clear();
    }
    void add(int from, int to, int cap, int cost) {
        edges.push_back(Edge(from, to, cap, 0, cost));
        edges.push_back(Edge(to, from, 0, 0, -cost));
        int m = edges.size();
        G[from].push_back(m - 2);
        G[to].push_back(m - 1);
    }
    bool SPFA(int s, int t, int &flow, int &cost) {
        for (int i = 0; i < N; i++) d[i] = INF;
        memset(inq, 0, sizeof(inq));
        d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;
        queue<int> que;
        que.push(s);
        while (!que.empty()) {
            int u = que.front();
            que.pop();
            inq[u]--;
            for (int i = 0; i < G[u].size(); i++) {
                Edge& e = edges[G[u][i]];
                if(e.cap > e.flow && d[e.to] > d[u] + e.cost) {
                    d[e.to] = d[u] + e.cost;
                    p[e.to] = G[u][i];
                    a[e.to] = min(a[u], e.cap - e.flow);
                    if(!inq[e.to]) {
                        inq[e.to]++;
                        que.push(e.to);
                    }
                }
            }
        }
        if(d[t] == INF) return false;
        flow += a[t];
        cost += d[t] * a[t];
        int u = t;
        while (u != s) {
            edges[p[u]].flow += a[t];
            edges[p[u]^1].flow -= a[t];
            u = edges[p[u]].from;
        }
        return true;
    }
    int MinMaxflow(int s, int t) {
        int flow = 0, cost = 0;
        while (SPFA(s, t, flow, cost));
        maxflow = flow; mincost = cost;
        return cost;
    }
};
int l[maxn], r[maxn], w[maxn], a[maxn];
int main()
{
    int n, m, s, t, k;
    scanf("%d%d", &n, &k);
    MCMF solve;
    int tot = 0;
    for (int i = 1; i <= n; i++) {
        int x1, x2, y1, y2;
        scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
        w[i] = sqrt((ll)(x1 - x2) * (x1 - x2) + (ll)(y1 - y2) * (y1 - y2));
        l[i] = x1<<1|1, r[i] = x2<<1;
        a[++tot] = l[i], a[++tot] = r[i];
        if(l[i] > r[i]) swap(l[i], r[i]);
    }
    sort(a + 1, a + 1 + tot);
    tot = unique(a + 1, a + 1 + tot) - a - 1;
    s = 0; t = 1001;
    solve.add(s, 1, k, 0); solve.add(tot, t, k, 0);
    for (int i = 1; i <= n; i++) {
        int x = lower_bound(a + 1, a + 1 + tot, l[i]) - a;
        int y = lower_bound(a + 1, a + 1 + tot, r[i]) - a;
        solve.add(x, y, 1, -w[i]);
    }
    for (int i = 1; i < tot; i++) solve.add(i, i + 1, INF, 0);
    printf("%d\n", -solve.MinMaxflow(s, t));
}