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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#790488 | #906. 强连通分量 | XY_Eleven | WA | 1ms | 3796kb | C++20 | 4.5kb | 2024-11-28 12:35:03 | 2024-11-28 12:35:04 |
Judging History
answer
#include <bits/stdc++.h>
// #include <windows.h>
#include <bits/extc++.h>
using namespace __gnu_pbds;
using namespace std;
//#pragma GCC optimize(3)
#define DB long double
#define LL long long
#define ULL unsigned long long
#define in128 __int128
#define cint const int
#define cLL const LL
#define For(z,e1,e2) for(int z=(e1);z<=(e2);z++)
#define Rof(z,e1,e2) for(int z=(e2);z>=(e1);z--)
#define For_(z,e1,e2) for(int z=(e1);z<(e2);z++)
#define Rof_(z,e1,e2) for(int z=(e2);z>(e1);z--)
#define inint(e) scanf("%d",&e)
#define inll(e) scanf("%lld",&e)
#define inpr(e1,e2) scanf("%d%d",&e1,&e2)
#define in3(e1,e2,e3) scanf("%d%d%d",&e1,&e2,&e3)
#define outint(e) printf("%d\n",e)
#define outint_(e) printf("%d%c",e," \n"[i==n])
#define outint2_(e,e1,e2) printf("%d%c",e," \n"[(e1)==(e2)])
#define outll(e) printf("%lld\n",e)
#define outll_(e) printf("%lld%c",e," \n"[i==n])
#define outll2_(e,e1,e2) printf("%lld%c",e," \n"[(e1)==(e2)])
#define exc(e) if(e) continue
#define stop(e) if(e) break
#define ret(e) if(e) return
#define ll(e) (1ll*(e))
#define pb push_back
#define ft first
#define sc second
#define pii pair<int,int>
#define pli pair<long long,int>
#define vct vector
#define clean(e) while(!e.empty()) e.pop()
#define all(ev) ev.begin(),ev.end()
#define sz(ev) ((int)ev.size())
#define debug(x) printf("%s=%d\n",#x,x)
#define x0 __xx00__
#define y1 __yy11__
#define ffo fflush(stdout)
cLL mod=998244353,G=404;
// cLL mod[2]={1686688681ll,1666888681ll},base[2]={166686661ll,188868881ll};
template <typename Type> void get_min(Type &w1,const Type w2) { if(w2<w1) w1=w2; } template <typename Type> void get_max(Type &w1,const Type w2) { if(w2>w1) w1=w2; }
template <typename Type> Type up_div(Type w1,Type w2) { return (w1/w2+(w1%w2?1:0)); }
template <typename Type> Type gcd(Type X_,Type Y_) { Type R_=X_%Y_; while(R_) { X_=Y_; Y_=R_; R_=X_%Y_; } return Y_; } template <typename Type> Type lcm(Type X_,Type Y_) { return (X_/gcd(X_,Y_)*Y_); }
template <typename Type> Type md(Type w1,const Type w2=mod) { w1%=w2; if(w1<0) w1+=w2; return w1; } template <typename Type> Type md_(Type w1,const Type w2=mod) { w1%=w2; if(w1<=0) w1+=w2; return w1; }
void ex_gcd(LL &X_,LL &Y_,LL A_,LL B_) { if(!B_) { X_=1ll; Y_=0ll; return ; } ex_gcd(Y_,X_,B_,A_%B_); X_=md(X_,B_); Y_=(1ll-X_*A_)/B_; } LL inv(LL A_,LL B_=mod) { LL X_=0ll,Y_=0ll; ex_gcd(X_,Y_,A_,B_); return X_; }
template <typename Type> void add(Type &w1,const Type w2,const Type M_=mod) { w1=md(w1+w2,M_); } void mul(LL &w1,cLL w2,cLL M_=mod) { w1=md(w1*md(w2,M_),M_); } template <typename Type> Type pw(Type X_,Type Y_,Type M_=mod) { Type S_=1; while(Y_) { if(Y_&1) mul(S_,X_,M_); Y_>>=1; mul(X_,X_,M_); } return S_; }
template <typename Type> Type bk(vector <Type> &V_) { auto T_=V_.back(); V_.pop_back(); return T_; } template <typename Type> Type tp(stack <Type> &V_) { auto T_=V_.top(); V_.pop(); return T_; } template <typename Type> Type frt(queue <Type> &V_) { auto T_=V_.front(); V_.pop(); return T_; }
template <typename Type> Type bg(set <Type> &V_) { auto T_=*V_.begin(); V_.erase(V_.begin()); return T_; } template <typename Type> Type bk(set <Type> &V_) { auto T_=*prev(V_.end()); V_.erase(*prev(V_.end())); return T_; }
mt19937 gen(43); int rd() { return abs((int)gen()); }
int rnd(int l,int r) { return rd()%(r-l+1)+l; }
void main_init()
{
}
cint N=5.01e5;
int n,m;
vct <int> v[N];
int dfn[N],low[N],id;
bool ins[N]; stack <int> st;
vct <vct<int> > scc;
void dfs(int p)
{
dfn[p]=low[p]=++id;
st.push(p),ins[p]=true;
for(auto i:v[p])
{
if(!dfn[i])
{
dfs(i);
get_min(low[p],low[i]);
}
else if(ins[i])
get_min(low[p],dfn[i]);
}
ret(low[p]!=dfn[p]);
scc.pb({});
while(true)
{
int t=st.top();
st.pop(),ins[t]=false;
scc.back().pb(t);
stop(t==p);
}
}
void main_solve()
{
inpr(n,m);
while(m--)
{
int x,y; inpr(x,y); x++,y++;
v[x].pb(y);
}
id=0;
For(i,1,n) if(!dfn[i])
dfs(i);
outint(sz(scc));
for(auto &i:scc)
{
printf("%d",sz(i));
for(auto p:i) printf(" %d",p-1);
printf("\n");
}
}
int main()
{
// ios::sync_with_stdio(0); cin.tie(0);
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
// srand(time(NULL));
main_init();
// int _; inint(_); For(__,1,_) // T>1 ?
// printf("\n------------\n\n"),
main_solve();
return 0;
}
/*
*/
Details
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Test #1:
score: 0
Wrong Answer
time: 1ms
memory: 3796kb
input:
6 7 1 4 5 2 3 0 5 5 4 1 0 3 4 2
output:
4 2 3 0 1 2 2 4 1 1 5
result:
wrong answer 5 is later than 2, but there is an edge (5, 2)