題目鏈接:https://codeforc.es/contest/1217/problem/F
思路:沒有那個在線的話,就完全是一道裸題,而那個在線可以發現總共只有兩種可能,那麼完全可以當做離線的來做,加些判斷就行了
#pragma GCC optimize(3)
#include <unordered_map>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%lld",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define mod ll(1000)
#define pb push_back
#define eps 1e-8
#define lc d<<1
#define rc d<<1|1
#define Pll pair<ll,ll>
#define P pair<int,int>
#define pi acos(-1)
int fa[200008],sz[200008],n,m,last;
unordered_map<int,int>mp[200008];
struct ds{
int op,x,y;}a[200008];
int fid(int x) {return (fa[x]^x)?fid(fa[x]):x;}
struct as{
int l,r;
vector<P>g;
}tr[200008<<2];
void build(int d,int l,int r)
{
tr[d].l=l,tr[d].r=r;
if(l==r) return;
int mid=(l+r)>>1;
build(lc,l,mid);
build(rc,mid+1,r);
}
void add(int d,int l,int r,P pos)
{
if(tr[d].l==l&&tr[d].r==r) {tr[d].g.pb(pos);return;}
int mid=(tr[d].l+tr[d].r)>>1;
if(mid>=r) add(lc,l,r,pos);
else if(l>mid) add(rc,l,r,pos);
else add(lc,l,mid,pos),add(rc,mid+1,r,pos);
}
void query(int d)
{
vector<int>st;
for(auto it:tr[d].g)
{
if(!mp[it.first][it.second]) continue;
int fx=fid(it.first),fy=fid(it.second);
if(fx==fy) continue;
if(sz[fx]<sz[fy]) swap(fx,fy);
fa[fy]=fx,sz[fx]+=sz[fy];
st.pb(fy);
}
if(tr[d].l==tr[d].r)
{
int z=tr[d].l;
int x=(a[z].x-1+last)%n+1,y=(a[z].y-1+last)%n+1;
//cout<<z<<" "<<x<<" "<<y<<endl;
if(x>y) swap(x,y);
if(a[z].op==2) cout<<(last=(fid(x)==fid(y)));
else mp[x][y]^=1;
}
else query(lc),query(rc);
for(int i=st.size()-1;i>=0;i--) sz[fa[st[i]]]-=sz[st[i]],fa[st[i]]=st[i];
st.clear();
}
int main()
{
cin.tie(0);
cout.tie(0);
cin>>n>>m;
build(1,1,m+1);
FOR(i,1,n) fa[i]=i,sz[i]=1;
FOR(i,1,m)
{
si(a[i].op),si(a[i].x),si(a[i].y);
if(a[i].op==1)
{
FOR(j,0,1)
{
int x=(a[i].x-1+j)%n+1,y=(a[i].y-1+j)%n+1;
if(x>y) swap(x,y);
if(mp[x].count(y)&&mp[x][y]<=i) add(1,mp[x][y],i,P(x,y));
mp[x][y]=i+1;
}
}
}
FOR(i,1,n)
{
for(auto it:mp[i]) if(it.second!=m+1) add(1,it.second,m+1,P(i,it.first));
mp[i].clear();
}
query(1);puts("");
return 0;
}