2105: 增强型LCP

Description

Input

Output

对于每个Lcp(a,b)操作输出最长公共前缀

Sample Input

47
abab
L 13
A 1 ab
L 1 3
C 56 cb
L 1 3
D 1 2
L 1 3

Sample Output

2
4
2
0
 
 
不会在splay上求lcp，如果没有这个操作就是裸的splay操作了。。。
然而加上了就不会了...
大家说因为修改数少，所以暴力重建然后二分hash求lcp即可。。。
 1 #include<iostream>
 2 #include<cstdlib>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<algorithm>
 7 #include<string>
 8 #include<map>
 9 #include<queue>
10 #include<vector>
11 #include<set>
12 #define inf 1000000000
13 #define maxn 1000000+5
14 #define maxm 10000+5
15 #define eps 1e-10
16 #define ll long long
17 #define ull unsigned long long
18 #define base 13131
19 #define for0(i,n) for(int i=0;i<=(n);i++)
20 #define for1(i,n) for(int i=1;i<=(n);i++)
21 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
22 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
23 #define for4(i,x) for(int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go)
24 using namespace std;
25 int read(){
26     int x=0,f=1;char ch=getchar();
27     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
28     while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
29     return x*f;
30 }
31 int n,m,q;
32 ull a[maxn],b[maxn];
33 char ch[maxn];
34 string s;
35 void rebuild(){
36     n=s.length();
37     b[n-1]=s[n-1];
38     for3(i,n-2,0)b[i]=b[i+1]*base+(ull)s[i];
39 }
40 ull get(int x,int l){
41     return b[x]-b[x+l]*a[l];
42 }
43 int main(){
44     //freopen("input.txt","r",stdin);
45     //freopen("output.txt","w",stdout);
46     n=read();q=read();
47     scanf("%s",ch);s=ch;
48     a[0]=1;
49     for1(i,maxn-1)a[i]=a[i-1]*(ull)base;
50     rebuild();
51     while(q--){
52         scanf("%s",ch);
53         if(ch[0]=='L'){
54             int x=read()-1,y=read()-1,l=0,r=n-y;
55             while(l<=r){
56                 int mid=(l+r)>>1;
57                 if(get(x,mid)==get(y,mid))l=mid+1;
58                 else r=mid-1;
59             }
60             printf("%d\n",r);
61         }
62         else if(ch[0]=='A'){
63             int x=read()-1;
64             scanf("%s",ch);
65             s.insert(x,ch);
66             rebuild();
67         }
68         else if(ch[0]=='C'){
69             int x=read()-1,y=read()-1;
70             scanf("%s",ch);
71             for2(i,x,y)
72                 s[i]=ch[i-x];
73             rebuild();
74         }
75         else{
76             int x=read()-1,y=read()-1;
77             s.erase(x,y-x+1);
78             rebuild();
79         }
80     }
81     return 0;
82 }
View Code

 

posted @ 2016-07-09 18:43  HTWX  阅读(158)  评论(0)  编辑  收藏  举报