bzoj1014:[JSOI2008]火星人prefix

思路:插入、修改操作是splay的模型,然后询问的话就可以二分答案,然后再用splay去判,关键就在于怎么去判断。

可以用字符串hash,splay每个节点维护一个hash域,然后就可以定义一个进制去hash即可二分判断,hash值让其自然溢出即可。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
#define maxn 100005
#define p 37
 
int n,m;
unsigned int power[maxn];
char s[maxn],t[5];
 
struct splay_tree{
    int ch[maxn][2],fa[maxn],note[maxn],size[maxn],root,tot;
    unsigned int hash[maxn],val[maxn];
    void update(int x){
        if (!x) return;int ls=ch[x][0],rs=ch[x][1];
        size[x]=size[ls]+size[rs]+1;
        hash[x]=hash[ls]*power[size[rs]+1]+val[x]*power[size[rs]]+hash[rs];
    }
    void newnode(int &x,int v,int bo){val[x=++tot]=v,note[x]=bo;}
    void build(int &x,int l,int r,int bo){
        if (l>r) return;int mid=(l+r)>>1;
        newnode(x,(mid!=0&&mid!=n+1)?s[mid]-'a':0,bo);
        build(ch[x][0],l,mid-1,0),build(ch[x][1],mid+1,r,1);
        fa[ch[x][0]]=fa[ch[x][1]]=x;
        update(x);
    }
    void rotate(int x){
        int y=fa[x],z=fa[y],bo=note[x],bo1=note[y];
        ch[y][bo]=ch[x][bo^1],fa[ch[x][bo^1]]=y;
        ch[x][bo^1]=y,fa[y]=x;
        fa[x]=z;if (bo1!=2) ch[z][bo1]=x;
        note[x]=bo1,note[y]=bo^1,note[ch[y][bo]]=bo;
        update(y);
    }
    void splay(int x){
        while (note[x]!=2){
            if (note[x]==note[fa[x]]) rotate(fa[x]);
            rotate(x);
        }
        root=x,update(x);
    }
    int find(int pos){
        int x=root;
        while (1){
            if (size[ch[x][0]]+1==pos) return splay(x),x;
            else if (size[ch[x][0]]>=pos) x=ch[x][0];
            else pos-=size[ch[x][0]]+1,x=ch[x][1];
        }
    }
    int suc(int x){x=ch[x][1];while (ch[x][0]) x=ch[x][0];return x;}
    void insert(int x,int v){
        splay(x);int y=suc(x),z;newnode(z,v,0);
        fa[ch[x][1]]=0,note[ch[x][1]]=2,splay(y);
        fa[y]=x,note[y]=1,ch[x][1]=y;
        fa[z]=y,ch[y][0]=z,splay(z);
    }
    void change(int x,int v){val[x]=v;splay(x);}
    unsigned int gethash(int l,int len){
        int r=l+len-1,x=find(l),y=find(r+2);
        splay(x),fa[ch[x][1]]=0,note[ch[x][1]]=2,splay(y);
        note[y]=1,fa[y]=x,ch[x][1]=y;
        unsigned int ans=hash[ch[y][0]];
        return splay(y),ans;
    }
    bool check(int x,int y,int len){return gethash(x,len)==gethash(y,len);}
}S;
 
void solve(int x,int y){
    int l=1,r=min(S.tot-x-1,S.tot-y-1),ans=0;
    while (l<=r){
        int mid=(l+r)>>1;
        if (S.check(x,y,mid)) l=mid+1,ans=mid;
        else r=mid-1;
    }
    printf("%d\n",ans);
}
 
int main(){
    scanf("%s",s+1),scanf("%d",&m);power[0]=1,n=strlen(s+1);
    for (int i=1;i<=100001;i++) power[i]=power[i-1]*p;
    S.build(S.root,0,n+1,2);
    while (m--){
        int x,y;scanf("%s",t+1);
        if (t[1]=='I') scanf("%d%s",&x,t+1),S.insert(S.find(x+1),t[1]-'a');
        else if (t[1]=='R') scanf("%d%s",&x,t+1),S.change(S.find(x+1),t[1]-'a');
        else scanf("%d%d",&x,&y),solve(x,y);
    }
    return 0;
}

 

posted @ 2016-10-28 11:29  DUXT  阅读(203)  评论(0编辑  收藏  举报