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 laofu
const 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) ;
else
rotate(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*/