[BZOJ1014] [JSOI2008]火星人prefix

/*平衡树根节点维护子树区间HASH值*/
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#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(120)  评论(0编辑  收藏  举报