传送门:HDU - 6858
思路分析
因为有动态删边加边,所以肯定选择LCT
用双指针统计出每个左端点构成环最近的右端点,查询的时候看$ans[l]$是否小于等于$Qr$就行了,因为$ans$是单调递增的
AC代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
bool IOerror=0;
inline char nc() {
static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
if(p1==pend) {
p1=buf;
pend=buf+fread(buf,1,BUF_SIZE,stdin);
if(pend==p1) {
IOerror=1;
return -1;
}
}
return *p1++;
}
inline bool blank(char ch) {
return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
}
template<class T> inline bool read(T &x) {
bool sign=0;
char ch=nc();
x=0;
for(; blank(ch); ch=nc());
if(IOerror)return false;
if(ch=='-')sign=1,ch=nc();
for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
if(sign)x=-x;
return true;
}
template<class T,class... U>bool read(T& h,U&... t) {
return read(h)&&read(t...);
}
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int INF = 0x3f3f3f3f;
const int N = 1e6+10;
namespace LCT {
#define fa(x) (tree[x].fa)
#define ls(x) (tree[x].ch[0])
#define rs(x) (tree[x].ch[1])
#define ident(x,f) (rs(f)==x)
#define connect(x,f,s) tree[f].ch[s]=x,tree[x].fa=f
#define reverse(x) swap(ls(x),rs(x)),tree[x].tag^=1
#define notroot(x) (ls(fa(x)) == x || rs(fa(x)) == x)
struct node {
int fa,ch[2],val,res;
bool tag;
} tree[N];
inline void pp(int x) {
tree[x].res=tree[ls(x)].res^tree[rs(x)].res^tree[x].val;
}
inline void pushdown(int x) {
if(tree[x].tag) {
if(ls(x)) reverse(ls(x));
if(rs(x)) reverse(rs(x));
}
tree[x].tag=0;
}
inline void pushall(int x) {
if(notroot(x)) pushall(fa(x));
pushdown(x);
}
inline void rotate(int x) {
int f = fa(x),ff=fa(f),fs=ident(x,f),ffs=ident(f,ff);
connect(tree[x].ch[fs^1],f,fs);
fa(x) = ff;
if(notroot(f)) tree[ff].ch[ffs] = x;
connect(f,x,fs^1);
pp(f),pp(x);
}
inline void splaying(int x) {
pushall(x);
while(notroot(x)) {
int f=fa(x),ff=fa(f);
if(notroot(f)) ident(f,ff)^ident(x,f) ? rotate(x):rotate(f);
rotate(x);
}
}
inline void access(int x) {
for(int y=0; x; x=fa(x)) {
splaying(x);
rs(x)=y;
pp(x);
y=x;
}
}
inline void mkroot(int x) {
access(x);
splaying(x);
reverse(x);
}
inline int findroot(int x) {
access(x);
splaying(x);
while(ls(x)) {
pushdown(x);
x=ls(x);
}
splaying(x);
return x;
}
inline void link(int x,int y) {
mkroot(x);
if(findroot(y)==x) return;
fa(x)=y;
}
inline void cut(int x,int y) {
mkroot(x);
if(findroot(y)!=x||fa(y)!=x||ls(y)) return;
fa(y)=rs(x)=0;
pp(x);
}
inline void split(int x,int y) {
mkroot(x);
access(y);
splaying(y);
}
#undef fa
#undef ls
#undef rs
#undef ident
#undef reverse
#undef connect
#undef notroot
};
int ans[N];
pair<int,int>a[N];
int main() {
#ifdef xiaofan
freopen("1.in","r",stdin);
freopen("1.out","w",stdout);
#endif
int T;
read(T);
while(T--){
int n,m,q;
read(n,m,q);
for(int i=1;i<=m;i++) read(a[i].fi,a[i].se);
for(int i=1;i<=m;i++) ans[i] = INF;
int l=1,r=1;
while(r<=m){
while(LCT::findroot(a[r].fi) == LCT::findroot(a[r].se)){
ans[l] = r;
LCT::cut(a[l].fi,a[l].se);
l++;
}
LCT::link(a[r].fi,a[r].se);
++r;
}
for(int i = l;i<=m;i++) LCT::cut(a[i].fi,a[i].se);
int last = 0;
while(q--){
int l,r;
read(l,r);
l = (l^last)%m+1;
r = (r^last)%m+1;
if(l>r) swap(l,r);
if(ans[l]<=r) {
puts("Yes");
last = 1;
}else{
puts("No");
last = 0;
}
}
}
return 0;
}