Problem - E - Codeforces

思路：想到用图论前驱图了，但是因为考虑可能有环的存在，但是其实题干中说明了不能通过一种或几种混合得到自己，所以就保证了不存在环，那就能用拓扑结构的性质做，用记忆化搜索就可以了

// Problem: E. Nastya and Potions
// Contest: Codeforces - Codeforces Round 888 (Div. 3)
// URL: https://codeforces.com/contest/1851/problem/E
// Memory Limit: 256 MB
// Time Limit: 3000 ms#include<bits/stdc++.h>
#include<sstream>
#include<cassert>
#define fi first
#define se second
#define i128 __int128
using namespace std;
typedef long long ll;
typedef double db;
typedef pair<int,int> PII;
const double eps=1e-7;
const int N=5e5+7 ,M=5e5+7, INF=0x3f3f3f3f,mod=1e9+7,mod1=998244353;
const long long int llINF=0x3f3f3f3f3f3f3f3f;
inline ll read() {ll x=0,f=1;char c=getchar();while(c<'0'||c>'9') {if(c=='-') f=-1;c=getchar();}
while(c>='0'&&c<='9') {x=(ll)x*10+c-'0';c=getchar();} return x*f;}
inline void write(ll x) {if(x < 0) {putchar('-'); x = -x;}if(x >= 10) write(x / 10);putchar(x % 10 + '0');}
inline void write(ll x,char ch) {write(x);putchar(ch);}
void stin() {freopen("in_put.txt","r",stdin);freopen("my_out_put.txt","w",stdout);}
bool cmp0(int a,int b) {return a>b;}
template<typename T> T gcd(T a,T b) {return b==0?a:gcd(b,a%b);}
template<typename T> T lcm(T a,T b) {return a*b/gcd(a,b);}
void hack() {printf("\n----------------------------------\n");}int T,hackT;
int n,m,k;
ll cost[N];
bool st[N];
int h[N],e[M],ne[M],w[M],idx;void add(int a,int b) {e[idx]=b,ne[idx]=h[a],h[a]=idx++;
}void dfs(int u) {if(h[u]==-1) {st[u]=true;return ;}ll res=0;for(int i=h[u];i!=-1;i=ne[i]) {int j=e[i];if(!st[j]) dfs(j);res+=cost[j];}st[u]=true;cost[u]=min(res,cost[u]);
}void solve() {n=read(),k=read();for(int i=1;i<=n;i++) cost[i]=read(),st[i]=false;for(int i=1;i<=k;i++) {int c=read();cost[c]=0;st[c]=true;}memset(h,-1,sizeof(int)*(n+4));idx=0;for(int i=1;i<=n;i++) {int m=read();for(int j=1;j<=m;j++) {int c=read();add(i,c);}}for(int i=1;i<=n;i++) {if(!st[i]) dfs(i);printf("%d ",cost[i]);}printf("\n");
}   int main() {// init();// stin();// ios::sync_with_stdio(false); scanf("%d",&T);// T=1; while(T--) hackT++,solve();return 0;       
}