「网络流 24 题」太空飞行计划

Posted 2021-12-28 uid001

大意: $m$个实验, 每个实验只能进行一次, 第$i$实验需要的仪器集合$R_i$, 收益$p_i$. $n$种仪器, 第$i$种仪器花费$c_i$, 每种仪器可以多次使用. 求最大收益.

数据范围$n,m\le 50$

 

 

最大权闭合子图.

源点$S$向第$i$个实验连边权为$p_i$, 第$i$个实验向集合$R_i$中的仪器连边权$INF$, 仪器$i$向汇点$T$连边权$c_i$. 则$\sum p_i$减去$S->T$的最小割就为最大收益.

对于实验与$S$间的割的含义为不选这个实验, 对于仪器与$T$之间的割的含义为选这个仪器.

跑完最大流后仍与$S$连通的点即为最优方案.

#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, S, T;
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,T) 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 main() 
	cin>>m>>n;
	cin.ignore();
	S = n+m+1, T = S+1;
	int sum = 0;
	REP(i,1,m) 
		string s;
		getline(cin,s);
		stringstream ss(s);
		int k, p;
		ss>>p;
		add(S,i,p);
		sum += p;
		while (ss>>k) add(i,k+m,INF);
	
	REP(i,1,n) 
		int r;
		cin>>r;
		add(i+m,T,r);
	
	int x = dinic();
	REP(i,1,m) if (dep[i]!=INF) printf("%d ",i);hr;
	REP(i,m+1,n+m) if (dep[i]!=INF) printf("%d ",i-m);hr;
	printf("%d\n", sum-x);