이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
/*input
12 17 10
1 2
2 3
3 4
1 5
2 6
3 7
4 8
5 6
6 7
7 8
5 9
6 10
7 11
8 12
9 10
10 11
11 12
6 3 1 7 12
3 7 1 2 3 4 5 6 7
11 3 1 3 5
9 2 1 9
8 4 1 2 3 4
1 1 1
12 0
10 3 1 6 10
11 8 2 3 5 6 7 9 10 11
8 7 2 3 4 5 6 7 8
*/
#include<bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<long long, null_type, less_equal<long long>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
//order_of_key #of elements less than x
// find_by_order kth element
using ll = long long;
using ld = long double;
using pii = pair<int,int>;
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(int i=0;i<n;i++)
#define REP1(i,n) for(int i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
const ll INF64=4e18;
const int INF=0x3f3f3f3f;
const ll MOD=1e9+7;
const ld PI=acos(-1);
const ld eps=1e-9;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
ll mult(ll a,ll b){
return a*b%MOD;
}
ll mypow(ll a,ll b){
if(b<=0) return 1;
ll res=1LL;
while(b){
if(b&1) res=(res*a)%MOD;
a=(a*a)%MOD;
b>>=1;
}
return res;
}
const int maxn=1e5+5;
const ll maxlg=__lg(maxn)+2;
const int sq=400;
vector<int> v[maxn],rev[maxn];
int dp[maxn];
pii mx[maxn][sq+5];
bool in[maxn];
int main(){
ios::sync_with_stdio(false),cin.tie(0);
int n,m,q;
cin>>n>>m>>q;
REP(i,m){
int a,b;
cin>>a>>b;
--a,--b;
v[a].pb(b);
rev[b].pb(a);
}
REP(i,n) REP(j,sq) mx[i][j]={0,n};
REP(i,n){
REP(j,sq){
//cout<<i<<' '<<mx[i][j].f<<' '<<mx[i][j].s<<'\n';
if(mx[i][j].s==n){
mx[i][j]={0,i};
break;
}
}
for(auto x:v[i]){
int l=0,r=0;
vector<pii> ans;
while(sz(ans)<sq and (l<sq or r<sq)){
pii tmp={mx[i][r].f+1,mx[i][r].s};
if(mx[i][r].s==n and mx[x][l].s==n) break;
if(mx[i][r].s==n) tmp={-1,-1};
if(r==sq or mx[x][l]>tmp){
if(!in[mx[x][l].s]) ans.pb(mx[x][l]);
++l;
}
else{
if(!in[mx[i][r].s]) ans.pb(tmp);
++r;
}
in[ans.back().s]=1;
}
REP(j,sz(ans)) mx[x][j]=ans[j],in[ans[j].s]=0;
}
}
in[n]=1;
while(q--){
int x,y;
cin>>x>>y;
--x;
if(y<sq){
vector<int> v;
REP(i,y){
int z;
cin>>z;
--z;
v.pb(z),in[z]=1;
}
int ans=-1;
for(auto z:mx[x]){
if(!in[z.s]) MXTO(ans,z.f);
}
for(auto x:v) in[x]=0;
cout<<ans<<'\n';
}
else{
REP(i,n) dp[i]=-INF;
REP(i,y){
int z;
cin>>z;
--z;
in[z]=1;
}
dp[x]=0;
for(int i=x;i>=0;i--){
for(auto z:rev[i]){
MXTO(dp[z],dp[i]+1);
}
}
int ans=-1;
REP(i,x+1) if(!in[i]) MXTO(ans,dp[i]);
REP(i,n) in[i]=0;
cout<<ans<<'\n';
}
}
}
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