最长k可重区间集问题

费用流，离散化后，l向r连费用为负长度的边容量为1
相邻的连容量为k的边，最好建S和T

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
# define Copy(a, b) memcpy(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1010), __(1e6 + 10), INF(2e9);

IL ll Read(){
    char c = '%'; ll x = 0, z = 1;
    for(; c > '9' || c < '0'; c = getchar()) if(c == '-') z = -1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
    return x * z;
}

int n, k, l[_], r[_], o[_], len;
int cnt, fst[_], w[__], to[__], nxt[__], dis[_], vis[_], S, T, cost[__], pe[_], pv[_], max_flow, max_cost;
queue <int> Q;

IL void Add(RG int u, RG int v, RG int f, RG int co){
    cost[cnt] = co; w[cnt] = f; to[cnt] = v; nxt[cnt] = fst[u]; fst[u] = cnt++;
    cost[cnt] = -co; w[cnt] = 0; to[cnt] = u; nxt[cnt] = fst[v]; fst[v] = cnt++;
}

IL bool Bfs(){
    Q.push(S); Fill(dis, 127); dis[S] = 0; vis[S] = 1;
    while(!Q.empty()){
        RG int u = Q.front(); Q.pop();
        for(RG int e = fst[u]; e != -1; e = nxt[e]){
            if(!w[e] || dis[to[e]] <= dis[u] + cost[e]) continue;
            dis[to[e]] = dis[u] + cost[e];
            pe[to[e]] = e; pv[to[e]] = u;
            if(!vis[to[e]]) vis[to[e]] = 1, Q.push(to[e]);
        }
        vis[u] = 0;
    }
    if(dis[T] >= dis[T + 1]) return 0;
    RG int ret = INF;
    for(RG int u = T; u != S; u = pv[u]) ret = min(ret, w[pe[u]]);
    for(RG int u = T; u != S; u = pv[u]) w[pe[u]] -= ret, w[pe[u] ^ 1] += ret;
    max_cost -= ret * dis[T]; max_flow += ret;
    return 1;
}

int main(RG int argc, RG char *argv[]){
    Fill(fst, -1); n = Read(); k = Read();
    for(RG int i = 1; i <= n; ++i){
        l[i] = Read(); r[i] = Read();
        o[++cnt] = l[i]; o[++cnt] = r[i];
    }
    sort(o + 1, o + cnt + 1); len = unique(o + 1, o + cnt + 1) - o - 1;
    cnt = 0; T = len + 1;
    for(RG int i = 0; i <= len; ++i) Add(i, i + 1, k, 0);
    for(RG int i = 1; i <= n; ++i){
        if(l[i] >= r[i]) continue;
        RG int le = r[i] - l[i];
        l[i] = lower_bound(o + 1, o + len + 1, l[i]) - o;
        r[i] = lower_bound(o + 1, o + len + 1, r[i]) - o;
        Add(l[i], r[i], 1, -le);
    }
    while(Bfs()); printf("%d\n", max_cost);
    return 0;
}