这题的写法非常巧妙。

每个位置的点向它的下一个位置连一个容量为\(INF\)的边，从区间的左端点往右端点拉一条容量为\(1\)，费用为区间长度的边，从起点向点\(1\)连一条容量为\(k\)的边，限制只能流\(k\)次。这个建模的巧妙之处就在，它并没有明面上限制某个点最多经过\(k\)次，却实际上使路径单向，从而每一次流过都最多经过每个点一次，同时也并不会对并没有相交的区间造成影响。

#include 
    
     using namespace std;const int N = 100010;const int M = 400010;const int INF = 0x3f3f3f3f;int n, k, tot, l[N], r[N], len[N], pos[N];int cnt = -1, head[N];struct edge {    int nxt, to, w, f;}e[M];void add_edge (int from, int to, int flw, int val) {    e[++cnt].nxt = head[from];    e[cnt].to = to;    e[cnt].w = val;    e[cnt].f = flw;    head[from] = cnt;}void add_len (int u, int v, int f, int w) {    add_edge (u, v, f, +w);    add_edge (v, u, 0, -w);}void discretize () {    for (int i = 1; i <= n; ++i) {        cin >> l[i] >> r[i];        len[i] = r[i] - l[i];        pos[i * 2 - 1] = l[i], pos[i * 2] = r[i];    }    sort (pos + 1, pos + 1 + n * 2);    tot = unique (pos + 1, pos + 1 + n * 2) - pos - 1;    for (int i = 1; i <= n; ++i) {        l[i] = lower_bound (pos + 1, pos + 1 + tot, l[i]) - pos;        r[i] = lower_bound (pos + 1, pos + 1 + tot, r[i]) - pos;    }}queue 
     
       q;int dis[N], vis[N], flow[N];int pre_node[N], pre_edge[N];int spfa (int s, int t) {    memset (vis, 0, sizeof (vis));    memset (dis, -0x3f, sizeof (dis));    memset (flow, 0x3f, sizeof (flow));    q.push (s); vis[s] = true; dis[s] = 0;    while (!q.empty ()) {        int u = q.front (); q.pop ();        for (int i = head[u]; ~i; i = e[i].nxt) {            int v = e[i].to;            if (dis[v] < dis[u] + e[i].w && e[i].f) {                dis[v] = dis[u] + e[i].w;                flow[v] = min (flow[u], e[i].f);                pre_edge[v] = i;                pre_node[v] = u;                if (!vis[v]) {                    vis[v] = true;                    q.push (v);                }            }        }        vis[u] = false;    }    return dis[t] != dis[0];}int main () {    memset (head, -1, sizeof (head));    cin >> n >> k;    discretize ();    int s = tot + 1, t = tot + 2;    add_len (s, 1, k, 0);    add_len (tot, t, k, 0);    for (int i = 1; i < tot; ++i) {        add_len (i, i + 1, INF, 0);    }    for (int i = 1; i <= n; ++i) {        add_len (l[i], r[i], 1, len[i]);    }    int max_cost = 0;    while (spfa (s, t)) {        max_cost += dis[t] * flow[t];        int u = t;        while (u != s) {            e[pre_edge[u] ^ 0].f -= flow[t];            e[pre_edge[u] ^ 1].f += flow[t];            u = pre_node[u];        }    }    cout << max_cost << endl;}