二分答案以后就是个最小路径覆盖。
注意要先Floyd传递闭包。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define N 510
inline char gc(){
static char buf[1<<16],*S,*T;
if(S==T){T=(S=buf)+fread(buf,1,1<<16,stdin);if(T==S) return EOF;}
return *S++;
}
inline int read(){
int x=0,f=1;char ch=gc();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=gc();}
while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=gc();
return x*f;
}
int K,n,aa[N],v[N],a[N],h[N],num=0,bf[N];
bool mp[N][N],f[N];
struct edge{
int to,next;
}data[N*N];
inline void add(int x,int y){
data[++num].to=y;data[num].next=h[x];h[x]=num;
}
inline bool find(int x){
for(int i=h[x];i;i=data[i].next){
int y=data[i].to;if(f[y]) continue;f[y]=1;
if(!bf[y]||find(bf[y])){bf[y]=x;return 1;}
}return 0;
}
inline bool jud(int val){
int tot=0;memset(h,0,sizeof(h));num=0;memset(bf,0,sizeof(bf));
for(int i=1;i<=n;++i) if(v[i]<=val) a[++tot]=i;int res=tot;
if(res<=K) return 1;
for(int i=1;i<=tot;++i)
for(int j=1;j<=tot;++j)
if(mp[a[i]][a[j]]) add(i,j);
for(int i=1;i<=tot;++i){
memset(f,0,sizeof(f));if(find(i)) --res;
}return res<=K;
}
int main(){
// freopen("contest.in","r",stdin);
// freopen("contest.out","w",stdout);
K=read()+1;n=read();
for(int i=1;i<=n;++i){
aa[i]=v[i]=read();int owo=read();
while(owo--) mp[i][read()]=1;
}sort(aa+1,aa+n+1);int mm=unique(aa+1,aa+n+1)-aa-1;
for(int k=1;k<=n;++k)
for(int i=1;i<=n;++i){
if(!mp[i][k]) continue;
for(int j=1;j<=n;++j)
mp[i][j]|=mp[k][j];
}
int l=1,r=mm;
while(l<=r){
int mid=l+r>>1;
if(jud(aa[mid])) l=mid+1;
else r=mid-1;
}if(l>mm) puts("AK");
else printf("%d\n",aa[l]);
return 0;
}