BZOJ 3237([Ahoi2013]连通图-cdq图重构-连通性缩点)

 

3237: [Ahoi2013]连通图

Time Limit: 20 Sec   Memory Limit: 512 MB
Submit: 106   Solved: 31
[ Submit][ Status]

Description

Input

Output

Sample Input

4 5
1 2
2 3
3 4
4 1
2 4
3
1 5
2 2 3
2 1 2

Sample Output

Connected
Disconnected
Connected

HINT

 

 

N<=100000 M<=200000 K<=100000

 

Source

 

 

 

弱B。。的弱B题解。。。

首先我们知道，可以把提问中没问的边缩成点。

但是不影响复杂度。。。

所以我们，把它拆成2半。。

前一半缩点(不考虑后一半的询问)，乱搞，后一半的不用考虑前一半的询问，乱搞。。。

于是f(q)=f(q/2)+O(qc*a(qc)) O(f(q))=O(qlogqc*α(qc)) 

 

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEMr(a,n,w) Rep(i,n) a[i]=w; 
#define MEMF(a,n,w) For(i,n) a[i]=w; 
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (100000+10)
#define MAXM (200000+10)
#define MAXQ (100000+10)
#define MAXC (4)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
int n,m,q;
struct comm
{
	int n,a[4];
}ask[MAXQ],back[MAXQ*30],*back_tail=back;
struct E
{
	int x,y;
}e[MAXM*30],*e_tail=e;
struct unionset
{
	int father[MAXN];
	void init(int n){For(i,n) father[i]=i;}
	int getfather(int x)
	{
		if (father[x]==x) return x;
		return father[x]=getfather(father[x]);
	}
	bool union2(int x,int y)
	{
		if (getfather(x)==getfather(y)) return 0;
		father[father[x]]=father[y]; return 1;
	}
}ufs;

bool ans[MAXQ]={0};
int newV[MAXN],newE[MAXM];
void solve(int n,E *_e,int m,int l,int r)
{
	e_tail+=m;
	E *e=e_tail;
	copy(_e,e_tail,e);
	static bool b[MAXM]={0};MEMr(b,m,0);
	if (l==r)
	{
		Rep(j,ask[l].n) b[ask[l].a[j]]=1;
		ufs.init(n);
		int tot=0;
		Rep(i,m) if (!b[i]) tot+=ufs.union2(e[i].x,e[i].y);
		if (tot==n-1) ans[l]=1;
		e_tail-=m;
		return;
	}
	Fork(i,l,r) Rep(j,ask[i].n) b[ask[i].a[j]]=1;
	ufs.init(n);
	Rep(i,m) if (!b[i]) ufs.union2(e[i].x,e[i].y);
	//Con
	int n2=0;
	For(i,n) if (ufs.getfather(i)==i) newV[i]=++n2;
	For(i,n) if (ufs.getfather(i)^i) newV[i]=newV[ufs.getfather(i)];
	Rep(i,m) e[i].x=newV[e[i].x],e[i].y=newV[e[i].y];
	//Red
	int m2=0;
	Rep(i,m) if (b[i]) newE[i]=m2++;
	Rep(i,m) if (b[i]) e[newE[i]]=e[i];
	Fork(i,l,r) Rep(j,ask[i].n) ask[i].a[j]=newE[ask[i].a[j]];
	
	{
		int m=l+r>>1,len=m-l+1;
		comm *back_head=back_tail;
		back_tail+=len;
		copy(ask+l,ask+m+1,back_head);
		solve(n2,e,m2,l,m);
		copy(back_head,back_head+len,ask+l);
		back_tail-=len;
		solve(n2,e,m2,m+1,r);		
	}	
	e_tail-=m;
	
}
int main()
{
//	freopen("bzoj3237.in","r",stdin);
	scanf("%d%d",&n,&m);
	Rep(i,m) scanf("%d%d",&e[i].x,&e[i].y);
	scanf("%d",&q);
	Rep(i,q)
	{
		scanf("%d",&ask[i].n);
		Rep(j,ask[i].n) scanf("%d",&ask[i].a[j]),ask[i].a[j]--;
	}
	solve(n,e,m,0,q-1);
	Rep(i,q) if (ans[i]) puts("Connected");else puts("Disconnected");
	return 0;
}