BZOJ 3495: PA2010 Riddle

/**************************************************************
    Problem: 3495
    User: BeiYu
    Language: C++
    Result: Accepted
    Time:20436 ms
    Memory:350776 kb
****************************************************************/
 
#include <bits/stdc++.h>
using namespace std;
 
const int N = 4000050;
const int M = 20000050;
 
inline int in(int x=0,char c=getchar()) { while(c>‘9‘||c<‘0‘) c=getchar();
    while(c>=‘0‘&&c<=‘9‘) x=x*10+c-‘0‘,c=getchar();return x; }
 
int n,k,m,cp,bp,tp;
int a[N];
int b[N],bl[N],dfn[N],low[N],stk[N];
 
int ep;
int nxt[M],tto[M],hd[M];
 
void AddEdge(int fr,int to) { nxt[++ep]=hd[fr],tto[ep]=to,hd[fr]=ep; }
void Tarjan(int u,int fa) {
    dfn[u]=low[u]=++cp,b[u]=1,stk[++tp]=u;
    for(int i=hd[u],v;i;i=nxt[i]) {
        v=tto[i];
        if(!dfn[v]) Tarjan(v,u),low[u]=min(low[u],low[v]);
        else if(b[v]) low[u]=min(low[u],dfn[v]);
    }
    if(low[u]==dfn[u]) {
        for(bp++;;) {
            int v=stk[tp];
            bl[v]=bp,b[v]=0,tp--;
            if(u==v) break;
        }
    }
}
int main() {
    n=in(),m=in(),k=in();
    for(int i=1;i<=m;i++) {
        int u=in(),v=in();
        AddEdge(u,v+n);
        AddEdge(v,u+n);
    }
    for(int i=1,t,j;i<=k;i++) {
        for(t=in(),j=1;j<=t;j++) a[j]=in();
        for(j=1;j<=t;j++) {
            if(j+1<=t)
                AddEdge(a[j+1]+2*n,a[j]+2*n),AddEdge(a[j]+3*n,a[j+1]+3*n),
                AddEdge(a[j]+3*n,a[j+1]),AddEdge(a[j+1]+n,a[j]+2*n);
            AddEdge(a[j]+n,a[j]+3*n),AddEdge(a[j]+2*n,a[j]);
        }
    }
    for(int i=1;i<=4*n;i++) if(!dfn[i]) Tarjan(i,i);
 
/*  for(int i=1;i<=4*n;i++) {
        cout<<i<<" --> ";
        for(int j=0;j<(int)g[i].size();j++) cout<<g[i][j]<<" ";cout<<endl;
    }
    for(int i=1;i<=4*n;i++) cout<<dfn[i]<<" ";cout<<endl;
    for(int i=1;i<=4*n;i++) cout<<low[i]<<" ";cout<<endl;
    for(int i=1;i<=4*n;i++) cout<<bl[i]<<" ";cout<<endl;*/
    for(int i=1;i<=n;i++) if(bl[i]==bl[i+n]) return puts("NIE"),0;
    for(int i=2*n+1;i<=3*n;i++) if(bl[i]==bl[i+n]) return puts("NIE"),0;
    return puts("TAK"),0;
}