BZOJ3495: PA2010 Riddle

$n leq 1e6,m leq 1e6$的无向图，每个点属于一个国家（我就说国家咋地），现要求给所有$k$个国家分别选首都，满足：每条边至少一个端点是首都；每个国家有且只有一个首都。

2-SAT。每个点选和不选。条件一直接满足，条件二边数太多。

前缀优化建边。来源

//#include<iostream>
#include<cstring>
#include<cstdio>
//#include<time.h>
//#include<complex>
//#include<set>
//#include<queue>
#include<algorithm>
#include<stdlib.h>
using namespace std;

#define LL long long
int qread()
{
    char c; int s=0; while ((c=getchar())<'0' || c>'9');
    do s=s*10+c-'0'; while ((c=getchar())>='0' && c<='9'); return s;
}

//Pay attention to '-' , LL and double of qread!!!!

int n,m,K;
#define maxn 4000011
#define maxm 10000011
struct Edge{int to,next;};
struct Graph
{
    Edge edge[maxm]; int first[maxn],le;
    Graph() {le=2;}
    void in(int x,int y) {Edge &e=edge[le]; e.to=y; e.next=first[x]; first[x]=le++;}
    void insert(int x,int y) {in(x,y); in(y,x);}
}g;

int bel[maxn],tot,Time,low[maxn],dfn[maxn],sta[maxn],top; bool insta[maxn];
void tarjan(int x)
{
    low[x]=dfn[x]=++Time;
    sta[++top]=x; insta[x]=1;
    for (int i=g.first[x];i;i=g.edge[i].next)
    {
        Edge &e=g.edge[i];
        if (!dfn[e.to]) tarjan(e.to),low[x]=min(low[x],low[e.to]);
        else if (insta[e.to]) low[x]=min(low[x],dfn[e.to]);
    }
    if (low[x]==dfn[x])
    {
        tot++;
        while (sta[top]!=x) insta[sta[top]]=0,bel[sta[top]]=tot,top--;
        insta[sta[top]]=0,bel[sta[top--]]=tot;
    }
}
bool check()
{
    for (int i=1;i<=n;i++) if (bel[i]==bel[i+n] || bel[i+2*n]==bel[i+3*n]) return 0;
    return 1;
}

int main()
{
    n=qread(); m=qread(); K=qread();
    for (int i=1,x,y;i<=m;i++)
    {
        x=qread(); y=qread();
        g.in(x+n,y); g.in(y+n,x);
    }
    for (int i=1,x,y,p;i<=K;i++)
    {
        x=qread(); p=0;
        for (int j=1;j<=x;j++)
        {
            y=qread();
            g.in(y,y+2*n); g.in(y+3*n,y+n);
            if (p)
            {
                g.in(p+2*n,y+2*n);
                g.in(y+3*n,p+3*n);
                g.in(p+2*n,y+n);
                g.in(y,p+3*n);
            }
            p=y;
        }
    }
    for (int i=1,to=4*n;i<=to;i++) if (!dfn[i]) tarjan(i);
    puts(check()?"TAK":"NIE");
    return 0;
}