【AHOI2013】【BZOJ3237】连通图

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

一开始以为是LCT,后来发现其实cdq+并查集就可做
其实LCT也可以.
考虑一个询问的影响范围,并不包括该询问前面的部分
允许离线,贡献可以累计,分治好了.
并查集维护,每次暴力把并查集的状态改回去.(其实用可持久化复杂度可能更优?

别轻信Description里的数据范围…看HINT…

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define MAXN 200010
#define SIZE 5000010
#define GET (ch>='0'&&ch<='9')
using namespace std;
inline void in(int &x)
{
    char ch=getchar();x=0;
    while (!GET)    ch=getchar();
    while (GET) x=x*10+ch-'0',ch=getchar();
}
int n,m,Q,tim;
int ans[MAXN],del[MAXN];
int f[MAXN],sta[SIZE],top;
int find(int x)
{
    if (f[x]==x)    return x;
    sta[++top]=x;sta[++top]=f[x];return f[x]=find(f[x]);
}
void Union(int a,int b) {   if (find(a)!=find(b))   sta[++top]=f[b],sta[++top]=f[f[b]],f[f[b]]=f[a];    }
struct edge {   int u,v;    }e[MAXN];
struct query    {   int c,id,e[4];  }q[MAXN];
void solve(int l,int r)
{
    int now=top,mid=(l+r)>>1,flag=1;
    if (l==r)
    {
        for (int i=0;i<q[l].c&&flag;i++)    if (find(e[q[l].e[i]].u)!=find(e[q[l].e[i]].v)) flag=0;
        ans[q[l].id]=flag;
        while (now!=top)    f[sta[top-1]]=sta[top],top-=2;
        return;
    }
    tim++;
    for (int i=l;i<=mid;i++)    for (int j=0;j<q[i].c;j++)  del[q[i].e[j]]=tim;
    for (int i=mid+1;i<=r;i++)  for (int j=0;j<q[i].c;j++)
        if (del[q[i].e[j]]!=tim)    Union(e[q[i].e[j]].u,e[q[i].e[j]].v);
    solve(l,mid);tim++;
    while (now!=top)    f[sta[top-1]]=sta[top],top-=2;
    for (int i=mid+1;i<=r;i++)  for (int j=0;j<q[i].c;j++)  del[q[i].e[j]]=tim;
    for (int i=l;i<=mid;i++)    for (int j=0;j<q[i].c;j++)
        if (del[q[i].e[j]]!=tim)    Union(e[q[i].e[j]].u,e[q[i].e[j]].v);
    solve(mid+1,r);
    while (now!=top)    f[sta[top-1]]=sta[top],top-=2;
}
int main()
{
    in(n);in(m);
    for (int i=1;i<=n;i++)  f[i]=i;
    for (int i=1;i<=m;i++)  in(e[i].u),in(e[i].v);
    in(Q);tim=1;
    for (int i=1;i<=Q;i++)
    {
        in(q[i].c);q[i].id=i;
        for (int j=0;j<q[i].c;j++)  in(q[i].e[j]),del[q[i].e[j]]=tim;
    }
    for (int i=1;i<=m;i++)  if (del[i]!=tim)    Union(e[i].u,e[i].v);
    solve(1,Q);
    for (int i=1;i<=Q;i++)  puts(ans[i]?"Connected":"Disconnected");
}