「网络流 24 题」星际转移

大意: $n$个空间站, 标号$1,...,n$, 地球在$0$, 月球在$-1$, $m$个飞船, 第$i$艘船每次只能载$h_i$人, 周期性停靠在集合$S_i$, 船每移动一次耗时为$1$. 初始$k$个人在地球, 求如何坐船能使$k$个人全到月球的时间最短.

记第$i$个空间站$t$时刻的状态为$(i,t)$.

$S$向$(0,0)$连边$k$, $(-1,t)$向$T$连边$INF$, $(i,t)$向$(i,t+1)$连边$INF$.

枚举时间, 每次将飞船的转移添进网络内, 若当前最大流为$k$, 则可以到达.

#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 '
'
#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



const int N = 1e6+10;
int n, m, k, S, T, tot;
int h[N], r[N], fa[N];
vector<int> g[N];
struct edge {
    int v,w,next;
} e[N];
int head[N], dep[N], vis[N], cur[N], cnt=1;
queue<int> Q;
void add(int u, int v, int w) {
    e[++cnt] = {v,w,head[u]};
    head[u] = cnt;
    e[++cnt] = {u,0,head[v]};
    head[v] = cnt;
}
int bfs() {
    REP(i,1,tot) dep[i]=INF,vis[i]=0,cur[i]=head[i];
    dep[S]=0,Q.push(S);
    while (Q.size()) {
        int u = Q.front(); Q.pop();
        for (int i=head[u]; i; i=e[i].next) {
            if (dep[e[i].v]>dep[u]+1&&e[i].w) {
                dep[e[i].v]=dep[u]+1;
                Q.push(e[i].v);
            }
        }
    }
    return dep[T]!=INF;
}
int dfs(int x, int w) {
    if (x==T) return w;
    int used = 0;
    for (int i=cur[x]; i; i=e[i].next) {
        cur[x] = i;
        if (dep[e[i].v]==dep[x]+1&&e[i].w) {
            int flow = dfs(e[i].v,min(w-used,e[i].w));
            if (flow) {
                used += flow;
                e[i].w -= flow;
                e[i^1].w += flow;
                if (used==w) break;
            }
        }
    }
    return used;
}
int dinic() {
    int ans = 0;
    while (bfs()) ans+=dfs(S,INF);
    return ans;
}

int Find(int x) {return fa[x]?fa[x]=Find(fa[x]):x;}
void add(int x, int y) {
	x = Find(x), y = Find(y);
	if (x!=y) fa[x] = y;
}
map<pii,int> s;
int id(int x, int y) {
	if (s.count(pii(x,y))) return s[pii(x,y)];
	return s[pii(x,y)]=++tot;
}

int main() {
	scanf("%d%d%d", &n, &m, &k);
	REP(i,1,m) {
		scanf("%d%d", h+i, r+i);
		REP(j,1,r[i]) {
			int x;
			scanf("%d", &x);
			g[i].pb(x);
			add(g[i].front()+2,g[i].back()+2);
		}
		if (r[i]==1) g[i].clear(),--i,--m;
	}
	if (Find(1)!=Find(2)) return puts("0"),0;
	int mf = 0;
	S = ++tot, T = ++tot;
	add(S,id(0,0),k);
	for (int t=1; ; ++t) {
		add(id(0,t-1),id(0,t),INF);
		REP(i,1,n) { 
			add(id(i,t-1),id(i,t),INF);
		}
		REP(i,1,m) {
			int L = id(g[i][(t%r[i]-1+r[i])%r[i]],t-1), R = id(g[i][t%r[i]],t);
			add(L,R,h[i]);
		}
		add(id(-1,t),T,INF);
		if ((mf+=dinic())==k) return printf("%d
",t),0;
	}
}