【BZOJ】1014: [JSOI2008]火星人prefix（splay+hash+二分+lcp）-白红宇

【BZOJ】1014: [JSOI2008]火星人prefix（splay+hash+二分+lcp）

阅读量：6091 次

发布时间：2019-06-20

本文共 3668 字，大约阅读时间需要 12 分钟。

题意：支持插入一个字符、修改一个字符，查询lcp。（总长度<=100000, 操作<=150000）

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             using namespace std;typedef long long ll;#define rep(i, n) for(int i=0; i<(n); ++i)#define for1(i,a,n) for(int i=(a);i<=(n);++i)#define for2(i,a,n) for(int i=(a);i<(n);++i)#define for3(i,a,n) for(int i=(a);i>=(n);--i)#define for4(i,a,n) for(int i=(a);i>(n);--i)#define CC(i,a) memset(i,a,sizeof(i))#define read(a) a=getint()#define print(a) printf("%d", a)#define dbg(x) cout << (#x) << " = " << (x) << endl#define error(x) (!(x)?puts("error"):0)#define rdm(x, i) for(int i=ihead[x]; i; i=e[i].next)inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }const int N=100005;const ll M=9875321;int P[N];struct node *null;struct node { node *f, *c[2]; int k, s, h; node(int _k=0) { k=_k; s=1; h=_k; f=c[0]=c[1]=null; } void setc(node *x, bool d) { c[d]=x; x->f=this; } bool d() { return f->c[1]==this; } void pushup() { s=c[0]->s+c[1]->s+1; h=(c[0]->h+((ll)k*P[c[0]->s])%M+((ll)c[1]->h*P[c[0]->s+1])%M)%M; }}*root;void rot(node *x) { node *f=x->f; bool d=x->d(); f->f->setc(x, f->d()); f->setc(x->c[!d], d); x->setc(f, !d); f->pushup(); if(f==root) root=x;}void splay(node *x, node *f=null) { while(x->f!=f) if(x->f->f==f) rot(x); else x->d()==x->f->d()?(rot(x->f), rot(x)):(rot(x), rot(x)); x->pushup();}node *sel(node *x, int k) { int s=x->c[0]->s; if(s==k) return x; if(s
              
               c[1], k-s-1); return sel(x->c[0], k);}node *getrange(int l, int r) { splay(sel(root, l-1)); splay(sel(root, r+1), root); return root->c[1];}void ins(int k, int pos) { node *f=getrange(pos+1, pos); f->setc(new node(k), 0); splay(f->c[0]);}void fix(int k, int pos) { node *f=getrange(pos, pos); f->c[0]->k=k; f->c[0]->pushup(); splay(f->c[0]);}int gethash(int l, int r) { return getrange(l, r)->c[0]->h; }int ask(int x, int y) { int s=root->s-2; int l=1, r=min(s-x, s-y)+1, mid; while(l<=r) { mid=(l+r)>>1; if(gethash(x, x+mid-1)==gethash(y, y+mid-1)) l=mid+1; else r=mid-1; } return l-1;}void U(node *x) { if(x==null) return; U(x->c[0]); U(x->c[1]); x->pushup(); }void Pr(node *x) { if(x==null) return; Pr(x->c[0]); printf("%c", x->k+'a'); Pr(x->c[1]); }void D(node *x=root) { Pr(x); puts(""); U(x); Pr(x); }char s[N];int n;void build(int l, int r, node *&x) { if(l>r) return; int mid=(l+r)>>1; x=new node(s[mid]-'a'); if(l==r) return; build(l, mid-1, x->c[0]); build(mid+1, r, x->c[1]); if(l<=mid-1) x->c[0]->f=x; if(mid+1<=r) x->c[1]->f=x; x->pushup();}void init() { P[0]=1; for1(i, 1, 100000) P[i]=(P[i-1]*26)%M; null=new node; null->s=0; null->f=null->c[0]=null->c[1]=null; root=new node; root->setc(new node, 1); node *x; build(1, n, x); root->c[1]->setc(x, 0); root->c[1]->pushup(); root->pushup(); //D();}char rdchar() { char ret=getchar(); while(ret<'a'||ret>'z') ret=getchar(); return ret;}int main() { scanf("%s", s+1); n=strlen(s+1); init(); int m=getint(); while(m--) { char c=getchar(); while(c!='Q'&&c!='R'&&c!='I') c=getchar(); if(c=='Q') { int x=getint(), y=getint(); printf("%d\n", ask(x, y)); } else if(c=='R') { int x=getint(); fix(rdchar()-'a', x); } else if(c=='I') { int x=getint(); ins(rdchar()-'a', x); } } return 0;}