BZOJ 1014: [JSOI2008]火星人prefix

Sol

Splay+Hash+二分答案.

用Splay维护Hash,二分答案判断.

复杂度 \(O(nlog^2n)\)

PS:这题调了两个晚上因为没开long long.许久不写数据结构题感觉写完整个人都不好了...

感觉还是应该经常开几道数据结构题来毒自己.

Code

/**************************************************************
    Problem: 1014
    User: BeiYu
    Language: C++
    Result: Accepted
    Time:6580 ms
    Memory:9320 kb
****************************************************************/
 
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
 
typedef unsigned long long LL;
const int N = 200050;
#define lc(o) d[o].ch[0]
#define rc(o) d[o].ch[1]
#define f(o) d[o].f
#define h(o) d[o].h
#define v(o) d[o].v
#define s(o) d[o].s
#define mid ((l+r)>>1)
 
//char *ps=(char *)malloc(N<<3);
inline int in(int x=0,char ch=getchar()){ while(ch>'9'||ch<'0') ch=getchar();
    while(ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();return x; }
 
LL p[N];char ch[N];
struct SplayTree{
    struct Node{
        int ch[2],f;
        LL h,v;int s;
        void Init(LL hh,LL vv,int ss){ h=hh,v=vv,s=ss,ch[0]=ch[1]=f=0; }
    }d[N];
    int rt,cnt;
     
    void PushUp(int o){
        if(!o) return;
        d[o].h=0;d[o].s=d[lc(o)].s+d[rc(o)].s+1;
        h(o)=(LL)h(lc(o))+v(o)*p[s(lc(o))]+h(rc(o))*p[s(lc(o))+1];
//      if(rc(o)!=0) h(o)=h(o)+h(rc(o));
//      h(o)=h(o)+v(o)*p[d[rc(o)].s];
//      if(lc(o)!=0) h(o)=h(o)+h(lc(o))*p[d[rc(o)].s+1];
    }
    void Build(int &o,int fa,int l,int r){
        if(l>r) return;
        o=++cnt;
        d[o].Init(0,0,0);
        if(ch[mid]>='a') d[o].v=ch[mid]-'a'+13;
        f(o)=fa;
        Build(lc(o),o,l,mid-1);
        Build(rc(o),o,mid+1,r);
        PushUp(o);
    }
//  int Build(int l,int r,int fa){
//      if(l==r){
//          d[l].Init(0,0,0);
//          if(ch[l]>='a') d[l].v=d[l].h=ch[l]-'a'+1;
//          d[l].s=1,f(l)=fa,PushUp(l);return l;
//      }
//      d[mid].Init(0,0,0);
//      if(ch[mid]>='a') v(mid)=ch[mid]-'a'+1;f(mid)=fa;
//      if(l<mid) lc(mid)=Build(l,mid-1,mid);
//      if(r>mid) rc(mid)=Build(mid+1,r,mid);
//      PushUp(mid);return mid;
//  }
    void Rot(int o){
        int p=f(o),k=f(p),r=rc(p)==o;
        if(k) d[k].ch[rc(k)==p]=o;
        d[d[o].ch[r^1]].f=p,d[o].f=k;
        d[p].ch[r]=d[o].ch[r^1],d[o].ch[r^1]=p,d[p].f=o;
        PushUp(p),PushUp(o);
    }
    void DFS(int o){
        if(lc(o)) DFS(lc(o));
        cout<<o<<" "<<d[o].v<<" "<<d[o].h<<" "<<d[o].s<<endl;
        cout<<" lc="<<lc(o)<<" rc="<<rc(o)<<" f="<<f(o)<<endl;
        if(rc(o)) DFS(rc(o));
    }
    void Splay(int o,int g){
        for(;f(o)!=g;){
            int p=f(o),k=f(p);
            if(k!=g) Rot((rc(p)==o)==(rc(k)==p)?p:o);
//          cout<<"Ok Rot"<<endl;
            Rot(o);
        }if(!g) rt=o;
//      cout<<"rt="<<rt<<endl;
//      DFS(rt);
    }
    int findkth(int o,int k){
        if(d[lc(o)].s>=k) return findkth(lc(o),k);
        else if(d[lc(o)].s+1<k) return findkth(rc(o),k-d[lc(o)].s-1);
        else return o;
    }
    void Insert(int x,LL v){
        int p=findkth(rt,x),q=findkth(rt,x+1);
        Splay(p,0),Splay(q,p);
//      d[++cnt].Init(0,v,0);rt=cnt;
//      rc(p)=0,f(p)=f(q)=rt,lc(rt)=p,rc(rt)=q;
//      PushUp(p),PushUp(q),PushUp(rt);
        d[++cnt].Init(0,v,0);
        f(cnt)=q,lc(q)=cnt;
        PushUp(cnt),PushUp(q),PushUp(p);
    }
    void Change(int x,LL v){
        int o=findkth(rt,x);
        Splay(o,0),d[o].v=v,PushUp(o);
    }
    void init(){
//      fread(ps,1,N<<3,stdin);
        p[0]=1;for(int i=1;i<N;i++) p[i]=(LL)p[i-1]*197; 
        d[0].Init(0,0,0);
        scanf("%s",ch+2);int n=strlen(ch+2);
        Build(rt,0,1,n+2);
//      rt=Build(1,n+2,0);
//      cout<<rt<<" "<<cnt<<endl;
    }
    LL Query(int u,int v){
//      if(u>v) return -1;
        u--,v++;
        u=findkth(rt,u),v=findkth(rt,v);
//      cout<<"Ok find "<<u<<" "<<v<<endl;
        Splay(u,0),Splay(v,u);
//      cout<<"Ok Splay"<<endl;
        return h(lc(v));
    }
    void sol(){
        char opt[5],c[5];int u,v;
        for(int m=in();m--;){
            scanf("%s",opt);
//          cout<<opt<<"********"<<endl;
            if(opt[0]=='R') scanf("%d%s",&u,c),Change(u+1,c[0]-'a'+13);
            else if(opt[0]=='I') scanf("%d%s",&u,c),Insert(u+1,c[0]-'a'+13);
            else {
                u=in()+1,v=in()+1;
                int l=0,r=min(cnt-v,cnt-u);
//              cout<<"Start Q"<<endl<<"*********"<<endl;
                while(l<=r){
//                  cout<<"***\ncs mid="<<mid<<endl;
//                  cout<<u<<" "<<mid<<" "<<Query(u,u+mid-1)<<endl;
//                  cout<<v<<" "<<mid<<" "<<Query(v,v+mid-1)<<endl;
                    if((LL)Query(u,u+mid-1)==Query(v,v+mid-1)) l=mid+1;
                    else r=mid-1;
                }printf("%d\n",r);
            }
//          cout<<opt<<" is ok"<<endl;
        }
    }
}spl;
 
int main(){
//  freopen("in.in","r",stdin);
//  freopen("out.out","w",stdout);
    spl.init();
//  cout<<"Finish init()"<<endl;
//  spl.DFS(spl.rt);
//  cout<<"Finish DFS()"<<endl;
    spl.sol();
//  cout<<"Finish sol()"<<endl;
    return 0;
}

  

posted @ 2016-09-19 20:52  北北北北屿  阅读(197)  评论(2编辑  收藏  举报