[BZOJ1014] [JSOI2008]火星人prefix 

/*平衡树根节点维护子树区间HASH值*/
  1 #include <cmath>
  2 #include <ctime>
  3 #include <cstdio>
  4 #include <cstring>
  5 #include <cstdlib>
  6 #include <iostream>
  7 #include <algorithm>
  8 # define maxn 100010
  9 # define p 66191
 10 using namespace std;
 11 typedef unsigned long long ULL;
 12 ULL px[maxn];
 13 void ot(){cout<<"***"<<endl;}
 14 void fen(){cout<<"-----------------"<<endl;}
 15 void beg(){px[0]=1; for(int i=1;i<=100000;i++) px[i]=px[i-1]*p;}
 16 struct Treap{
 17     Treap* ch[2];
 18     ULL hs1,hs2; int key,size,val; char lt;
 19     Treap(int v,char ss)
 20         {lt=ss; val=v;hs2=hs1=(ULL)ss; size=1; key=rand(); ch[0]=ch[1]=NULL;}
 21     ULL HS(Treap* now){
 22         return now?now->hs2:0;
 23     }
 24     int SZ(Treap *now){
 25         return now?now->size:0;
 26     }
 27     void update(){
 28         // cout<<HS(ch[0])<<"  "<<hs1<<"  "<<HS(ch[1])<<endl;
 29         size=1+(ch[0]?ch[0]->size:0)+(ch[1]?ch[1]->size:0);
 30         hs2=HS(ch[0])*px[SZ(ch[1])+1]+hs1*px[SZ(ch[1])]+HS(ch[1]);
 31     }
 32 }*root;
 33 void ot_t(Treap* now){
 34     if(!now) return;
 35     if(now->ch[0]) ot_t(now->ch[0]);
 36     cout<<now->lt;
 37     if(now->ch[1]) ot_t(now->ch[1]);
 38 }
 39 typedef pair<Treap*,Treap*> D;
 40 int Size(Treap* now){return now?now->size:0;}
 41 D Split(Treap* now,int k){
 42     if(!now) return D(NULL,NULL);
 43     D y;
 44     if(Size(now->ch[0])>=k)
 45         {y=Split(now->ch[0],k); now->ch[0]=y.second; now->update(); y.second=now;}
 46     else
 47         {y=Split(now->ch[1],k-Size(now->ch[0])-1); now->ch[1]=y.first; now->update(); y.first=now;}
 48     return y;
 49 }
 50 Treap* Merge(Treap *a,Treap* b){
 51     if(!a) return b;
 52     if(!b) return a;
 53     if(a->key<b->key)
 54         {a->ch[1]=Merge(a->ch[1],b); a->update(); return a;}
 55     else
 56         {b->ch[0]=Merge(a,b->ch[0]); b->update(); return b;}
 57 }
 58 void Insert(int pos,char ss){
 59     D x=Split(root,pos);
 60     Treap* now=new Treap(pos,ss);
 61     root=Merge(Merge(x.first,now),x.second);
 62 }
 63 void change(int pos,char ss){
 64     D x=Split(root,pos-1);
 65     D y=Split(x.second,1);
 66     Treap* now=new Treap(pos,ss);
 67     root=Merge(Merge(x.first,now),y.second);
 68 }
 69 char s[100010];
 70 int n;
 71 bool check(int a1,int a2,int mid){
 72     if(!mid) return 1;
 73     D x=Split(root,a1-1);
 74     D y=Split(x.second,mid);
 75     ULL now1=y.first->hs2;
 76     root=Merge(Merge(x.first,y.first),y.second);
 77     x=Split(root,a2-1);
 78     y=Split(x.second,mid);
 79     ULL now2=y.first->hs2;
 80     root=Merge(Merge(x.first,y.first),y.second);
 81     if(now1==now2) return 1;
 82     return 0;
 83 }
 84 void find_ans(int x,int y,int len){
 85     int l=0,r=len-y+1,mid,ans;
 86     while(l<=r){
 87         mid=(l+r)>>1;
 88         if(check(x,y,mid)) ans=mid,l=mid+1;
 89         else r=mid-1;
 90     }
 91     printf("%d\n",ans);
 92 }
 93 void work(){
 94     scanf("%s",&s);
 95     int len=strlen(s);
 96     ULL h=0;
 97     for(int i=0;i<len;i++){
 98         Insert(i,s[i]);
 99     }
100     scanf("%d",&n);
101     char od[3],ss; int x,y;
102     for(int i=1;i<=n;i++){
103         scanf("%s",&od);
104         if(od[0]=='Q'){
105             scanf("%d%d",&x,&y);
106             find_ans(x,y,len);
107         }
108         else if(od[0]=='I'){
109             scanf("%d%s",&x,&ss);
110             Insert(x,ss); len++;
111         }
112         else{
113             scanf("%d%s",&x,&ss);
114             change(x,ss);
115         }
116     }
117 }
118 int main(){
119     // freopen("a.in","r",stdin);
120     beg();
121     work();
122     return 0;
123 }

 

posted @ 2017-09-26 19:35  Nawox  阅读(149)  评论(0)    收藏  举报