CF13E Holes（分块儿LCT）

Posted 2022-10-18 oneman233

给你一堆弹簧，每个弹簧有一定的弹跳距离

现在把球放到任意一个弹簧上，问需要几次球跳过所有弹簧，并且输出最后一个跳出去的位置

还要支持更改操作

一开始乱胡并查集，但是不行，一旦路径压缩就死亡，直接暴力搞又T

rhy后来说是LCT裸体，维护每个点的深度和最靠近根节点的祖先就行，根节点编号不妨设为n+1，甚至是某一年OI原题

没想到分块儿也可以搞，把同一块内的暴力维护，块间直接跳，省去了在块内跳来跳去的时间

修改和查询均摊下来都是sqrt(n)的时间复杂度，但是还是加了快读快写卡过去

代码：

#include <bits/stdc++.h>
#define int long long
#define sc(a) scanf("%lld",&a)
#define scc(a,b) scanf("%lld %lld",&a,&b)
#define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define schar(a) scanf("%c",&a)
#define pr(a) printf("%lld",a)
#define fo(i,a,b) for(int i=a;i<b;++i)
#define re(i,a,b) for(int i=a;i<=b;++i)
#define rfo(i,a,b) for(int i=a;i>b;--i)
#define rre(i,a,b) for(int i=a;i>=b;--i)
#define prn() printf("\n")
#define prs() printf(" ")
#define mkp make_pair
#define pii pair<int,int>
#define pub(a) push_back(a)
#define pob() pop_back()
#define puf(a) push_front(a)
#define pof() pop_front()
#define fst first
#define snd second
#define frt front()
#define bak back()
#define mem0(a) memset(a,0,sizeof(a))
#define memmx(a) memset(a,0x3f3f,sizeof(a))
#define memmn(a) memset(a,-0x3f3f,sizeof(a))
#define debug
#define db double
#define yyes cout<<"YES"<<endl;
#define nno cout<<"NO"<<endl;
using namespace std;
typedef vector<int> vei;
typedef vector<pii> vep;
typedef map<int,int> mpii;
typedef map<char,int> mpci;
typedef map<string,int> mpsi;
typedef deque<int> deqi;
typedef deque<char> deqc;
typedef priority_queue<int> mxpq;
typedef priority_queue<int,vector<int>,greater<int> > mnpq;
typedef priority_queue<pii> mxpqii;
typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii;
const int maxn=500005;
const int inf=0x3f3f3f3f3f3f3f3f;
const int MOD=100000007;
const db eps=1e-10;
int qpow(int a,int b)int tmp=a%MOD,ans=1;while(b)if(b&1)ans*=tmp,ans%=MOD;tmp*=tmp,tmp%=MOD,b>>=1;return ans;
int lowbit(int x)return x&-x;
int max(int a,int b)return a>b?a:b;
int min(int a,int b)return a<b?a:b;
int mmax(int a,int b,int c)return max(a,max(b,c));
int mmin(int a,int b,int c)return min(a,min(b,c));
void mod(int &a)a+=MOD;a%=MOD;
bool chk(int now)
int half(int l,int r)while(l<=r)int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;return l;
int ll(int p)return p<<1;
int rr(int p)return p<<1|1;
int mm(int l,int r)return (l+r)/2;
int lg(int x)if(x==0) return 1;return (int)log2(x)+1;
bool smleql(db a,db b)if(a<b||fabs(a-b)<=eps)return true;return false;
db len(db a,db b,db c,db d)return sqrt((a-c)*(a-c)+(b-d)*(b-d));
bool isp(int x)if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;
inline int read()
    char ch=getchar();int s=0,w=1;
    while(ch<48||ch>57)if(ch==‘-‘)w=-1;ch=getchar();
    while(ch>=48&&ch<=57)s=(s<<1)+(s<<3)+ch-48;ch=getchar();
    return s*w;

inline void write(int x)
    if(x<0)putchar(‘-‘),x=-x;
    if(x>9)write(x/10);
    putchar(x%10+48);

 
int n,m,tot;
int a[maxn],cnt[maxn],pos[maxn];
int id[maxn],from[maxn],to[maxn];
int o,x,y;

void modify(int i)
    if(i+a[i]>n) pos[i]=i,cnt[i]=0;
    else
        if(id[i]==id[i+a[i]])
            pos[i]=pos[i+a[i]];
            cnt[i]=cnt[i+a[i]]+1;
        
        else
            pos[i]=i+a[i];
            cnt[i]=1;
        
    


void ask(int x)
    int p=x,res=0;
    while(p!=pos[p])
        res+=cnt[p],p=pos[p];
    write(p),printf(" "),write(res+1),puts("");


signed main()
    ios_base::sync_with_stdio(0);
    cin.tie(0),cout.tie(0);
    n=read(),m=read();
    tot=sqrt(n);
    re(i,1,tot)
        from[i]=(i-1)*tot+1;
        to[i]=i*tot;
    
    if(to[tot]<n) tot++,from[tot]=to[tot-1]+1,to[tot]=n;
    re(i,1,tot)
        re(j,from[i],to[i]) id[j]=i;
    
    re(i,1,n) a[i]=read();
    rre(i,n,1) modify(i);
    while(m--)
        o=read();
        if(o==0)
            x=read(),y=read();
            a[x]=y;
            rre(i,x,from[id[x]]) modify(i);
        
        else if(o==1) x=read(),ask(x);
    
    return 0;