http://acm.hdu.edu.cn/contests/contest_showproblem.php?pid=1004&cid=886
题意：
一张无向图,q个强制在线询问,如果只用区间[l,r]的边,能否组成一个简单环
解析：
求出每个i,至少要扩展到ans[i]才能使[i,ans[i]]组成简单环,ans[i]是单调的。
所以我们可以用尺取法来维护出ans[i]，往右延时的时候加边。如果当前边两点已经在一棵树上说明可以形成环。
缩短的时候需要删除边，所以使用动态树。
最后区间求一个最小值判断是否小于等于R即可。（ST表）
代码：
/** Author : Jk_Chen* Date : 2020-08-13-13.26.36*/
#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define LD long double
#define rep(i,a,b) for(int i=(int)(a);i<=(int)(b);i++)
#define per(i,a,b) for(int i=(int)(a);i>=(int)(b);i--)
#define mmm(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define pill pair<int, int>
#define fi first
#define se second
void test() {cerr<<"\n";
}
template<typename T,typename... Args>void test(T x,Args... args) {cerr<<"> "<<x<<" ";test(args...);
}
const LL mod=1e9+7;
const int maxn=3e5+9;
const int inf=0x3f3f3f3f;
LL rd() {LL ans=0;char last=' ',ch=getchar();while(!(ch>='0' && ch<='9'))last=ch,ch=getchar();while(ch>='0' && ch<='9')ans=ans*10+ch-'0',ch=getchar();if(last=='-')ans=-ans;return ans;
}
#define rd read()
/*_________________________________________________________begin*/namespace Fast_IO{ //orz laofuconst int MAXL((1 << 18) + 1);int iof, iotp;char ioif[MAXL], *ioiS, *ioiT, ioof[MAXL],*iooS=ioof,*iooT=ioof+MAXL-1,ioc,iost[55];char Getchar(){if (ioiS == ioiT){ioiS=ioif;ioiT=ioiS+fread(ioif,1,MAXL,stdin);return (ioiS == ioiT ? EOF : *ioiS++);}else return (*ioiS++);}void Write(){fwrite(ioof,1,iooS-ioof,stdout);iooS=ioof;}void Putchar(char x){*iooS++ = x;if (iooS == iooT)Write();}inline int read(){int x=0;for(iof=1,ioc=Getchar();(ioc<'0'||ioc>'9')&&ioc!=EOF;)iof=ioc=='-'?-1:1,ioc=Getchar();if(ioc==EOF)Write(),exit(0);for(x=0;ioc<='9'&&ioc>='0';ioc=Getchar())x=(x<<3)+(x<<1)+(ioc^48);return x*iof;}inline long long read_ll(){long long x=0;for(iof=1,ioc=Getchar();(ioc<'0'||ioc>'9')&&ioc!=EOF;)iof=ioc=='-'?-1:1,ioc=Getchar();if(ioc==EOF)Write(),exit(0);for(x=0;ioc<='9'&&ioc>='0';ioc=Getchar())x=(x<<3)+(x<<1)+(ioc^48);return x*iof;}void Getstr(char *s, int &l){for(ioc=Getchar();ioc==' '||ioc=='\n'||ioc=='\t';)ioc=Getchar();if(ioc==EOF)Write(),exit(0);for(l=0;!(ioc==' '||ioc=='\n'||ioc=='\t'||ioc==EOF);ioc=Getchar())s[l++]=ioc;s[l] = 0;}template <class Int>void Print(Int x, char ch = '\0'){if(!x)Putchar('0');if(x<0)Putchar('-'),x=-x;while(x)iost[++iotp]=x%10+'0',x/=10;while(iotp)Putchar(iost[iotp--]);if (ch)Putchar(ch);}void Putstr(const char *s){for(int i=0,n=strlen(s);i<n;++i)Putchar(s[i]);}
} // namespace Fast_IO
using namespace Fast_IO;namespace LCT {
struct node {int f, son[2] ;bool fz ;
} tr[maxn] ;inline void init(int n){rep(i,0,n)tr[i].f=tr[i].son[0]=tr[i].son[1]=tr[i].fz=0;
}inline void reverse ( int x ) {tr[x].fz = 0 ;swap ( tr[x].son[0], tr[x].son[1] ) ;int lc = tr[x].son[0] ;int rc = tr[x].son[1] ;tr[lc].fz ^= 1, tr[rc].fz ^= 1 ;
}inline void rotate ( int x, int w ) {int f = tr[x].f, ff = tr[f].f, r, R ;r = tr[x].son[w] ;R = f ;tr[R].son[1-w] = r ;if ( r )tr[r].f = R ;r = x ;R = ff ;if ( tr[R].son[0] == f )tr[R].son[0] = r ;else if ( tr[R].son[1] == f )tr[R].son[1] = r ;tr[r].f = R ;r = f ;R = x ;tr[R].son[w] = r ;tr[r].f = R ;
}int tmp[maxn] ;inline void splay ( int x, int rt ) {int s = 0, i = x ;while ( tr[i].f && (tr[tr[i].f].son[0]==i||tr[tr[i].f].son[1]==i) ) {tmp[++s] = i ;i = tr[i].f ;}tmp[++s] = i ;while ( s ) {i = tmp[s] ;s -- ;if ( tr[i].fz )reverse(i) ;}while ( tr[x].f!=rt && (tr[tr[x].f].son[0]==x||tr[tr[x].f].son[1]==x) ) {int f = tr[x].f, ff = tr[f].f ;if ( ff == rt || (tr[ff].son[0]!=f&&tr[ff].son[1]!=f) ) {if ( x == tr[f].son[0] )rotate(x,1) ;elserotate(x,0) ;} else {if ( tr[ff].son[0] == f && tr[f].son[0] == x )rotate(f,1), rotate(x,1) ;else if ( tr[ff].son[1] == f && tr[f].son[1] == x )rotate(f,0), rotate(x,0) ;else if ( tr[ff].son[0] == f && tr[f].son[1] == x )rotate(x,0), rotate(x,1) ;else if ( tr[ff].son[1] == f && tr[f].son[0] == x )rotate(x,1), rotate(x,0) ;}}
}inline void access ( int x ) {int y = 0 ;while ( x ) {splay ( x, 0 ) ;tr[x].son[1] = y ;if ( y != 0 )tr[y].f = x ;y = x ;x = tr[x].f ;}
}inline void makeroot ( int x ) {access(x) ;splay(x,0) ;tr[x].fz ^= 1 ;
}inline void link ( int x, int y ) {makeroot(x) ;tr[x].f = y ;access(x) ;
}inline void cut ( int x, int y ) {makeroot(x) ;access(y) ;splay(y,0) ;tr[tr[y].son[0]].f = 0 ;tr[y].son[0] = 0 ;
}int find_root ( int x ) {access(x) ;splay(x,0) ;while ( tr[x].son[0] != 0 )x = tr[x].son[0] ;return x ;
}
}
using namespace LCT;int n,m,q;
pill e[maxn];
int ans[maxn];
int len[maxn];
int mi[maxn][30];
void init_ST() {rep(i,1,maxn-1){len[i]=floor(log2(i));}int L=len[m];rep(i,1,m){mi[i][0]=ans[i];}rep(i,1,L) {rep(j,1,m){if(j+(1<<i)-1>m)break;mi[j][i]=min(mi[j][i-1],mi[j+(1<<i-1)][i-1]);}}
}
int findMin(int l,int r){int L=len[r-l+1];return min(mi[l][L],mi[r-(1<<L)+1][L]);
}int main() {int t=rd;while(t--){int last=0;n=rd,m=rd,q=rd;init(n);rep(i,1,m)e[i]={rd,rd};int l=1,r=0;while(r<m){r++;while(find_root(e[r].fi)==find_root(e[r].se)){ans[l]=r;cut(e[l].fi,e[l].se);l++;}link(e[r].fi,e[r].se);}rep(i,l,m)ans[i]=inf;init_ST();while(q--){int l=(rd^last)%m+1,r=(rd^last)%m+1;if(l>r)swap(l,r);int Left=findMin(l,r);if(Left<=r){puts("Yes");last=1;}else{puts("No");last=0;}}}return 0;
}/*_________________________________________________________end*/