BZOJ 3237([Ahoi2013]连通图-cdq图重构-连通性缩点)

来源：互联网发布：50元洗面奶推荐知乎编辑：程序博客网时间：2024/04/28 07:10

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);//Conint 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];//Redint 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;}