「网络流 24 题」最长 k 可重区间集

给定区间集合$I$和正整数$k$, 计算$I$的最长$k$可重区间集的长度.

区间离散化到$[1,2n]$, $S$与$1$连边$(k,0)$, $i$与$i+1$连边$(k,0)$, $2n$与$T$连边$(k,0)$. 对于每个区间$(l,r)$, $l$与$r$连边$(1,l-r)$.

最小费用相反数就为最大长度

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#include <unordered_map>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head
 
#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 999;
#endif
 
int n, m, k, S, T;
struct _ {int from,to,w,f;};
vector<_> E;
vector<int> g[N];
int a[N], pre[N], inq[N], d[N];
int mf,mc;
queue<int> q;
void add(int x, int y, int c, int w) {
    g[x].pb(E.size());
    E.pb({x,y,c,w});
    g[y].pb(E.size());
    E.pb({y,x,0,-w});
}
void mfmc() {
	mf=mc=0;
    while (1) {
        REP(i,1,T) a[i]=d[i]=INF,inq[i]=0;
        q.push(S),d[S]=0;
        while (!q.empty()) {
            int x=q.front(); q.pop();
            inq[x] = 0;
            for (auto t:g[x]) {
                auto e=E[t];
                if (e.w>0&&d[e.to]>d[x]+e.f) {
                    d[e.to]=d[x]+e.f;
                    pre[e.to]=t;
                    a[e.to]=min(a[x],e.w);
                    if (!inq[e.to]) {
                        inq[e.to]=1;
                        q.push(e.to);
                    }
                }
            }
        }
        if (a[T]==INF) break;
        for (int u=T;u!=S;u=E[pre[u]].from) {
            E[pre[u]].w-=a[T];
            E[pre[u]^1].w+=a[T];
        }
        mf+=a[T],mc+=a[T]*d[T];
    }
}
 
int b[N], l[N], r[N];
int main() {
	scanf("%d%d", &n, &k);
	REP(i,1,n) {
		scanf("%d%d",l+i,r+i);
		b[++*b]=l[i],b[++*b]=r[i];
	}
	sort(b+1,b+1+*b),*b=unique(b+1,b+1+*b)-b-1;
	REP(i,1,n) {
		l[i]=lower_bound(b+1,b+1+*b,l[i])-b;
		r[i]=lower_bound(b+1,b+1+*b,r[i])-b;
	}
	S = *b+1, T = S+1;
	add(S,1,k,0),add(*b,T,k,0);
	REP(i,2,*b) add(i-1,i,k,0);
	REP(i,1,n) add(l[i],r[i],1,-b[r[i]]+b[l[i]]);
	mfmc();
	printf("%d\n", -mc);
}