2105: 增强型LCP
Time Limit: 10 Sec Memory Limit: 162 MB
Submit: 366 Solved: 86
[Submit][Status]
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
HINT
Source
题解：
这题。。。
原来一直以为是个splay练手题，于是昨天来写。。。
2h没调出来，今天来了发现pushup写错了。。。T_T
然后就是狂T了。。。
看题解发现我们暴力重构写hash就行了，因为修改少，我也是醉了。。。
然后终于会了字符串的hash算法
我们另b[i]=si-sn的hash值，递推式b[i]=b[i+1]*base+s[i]
然后我们要得到从i开始的len长度的hash就是 b[i]-b[i+len]*a[len] a[len]表示base^len
还有c++的字符串问题
s,insert(pos,st) 表示在pos前插入st
s.erase(pos,len)表示从pos开始删除len的字符，包括pos
代码：
1 #include<cstdio>
2
3 #include<cstdlib>
4
5 #include<cmath>
6
7 #include<cstring>
8
9 #include<algorithm>
10
11 #include<iostream>
12
13 #include<vector>
14
15 #include<map>
16
17 #include<set>
18
19 #include<queue>
20
21 #include<string>
22
23 #define inf 1000000000
24
25 #define maxn 1000000+5
26
27 #define maxm 500+100
28
29 #define eps 1e-10
30
31 #define ull unsigned long long
32
33 #define pa pair<int,int>
34
35 #define for0(i,n) for(int i=0;i<=(n);i++)
36
37 #define for1(i,n) for(int i=1;i<=(n);i++)
38
39 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
40
41 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
42
43 #define mod 1000000007
44 #define base 13131
45
46 using namespace std;
47
48 inline int read()
49
50 {
51
52 int x=0,f=1;char ch=getchar();
53
54 while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
55
56 while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}
57
58 return x*f;
59
60 }
61 int n,m,q;
62 ull a[maxn],b[maxn];
63 char ch[maxn];
64 string s;
65 inline void rebuild()
66 {
67 n=s.length();
68 b[n-1]=s[n-1];
69 for3(i,n-2,0)b[i]=b[i+1]*base+(ull)s[i];
70 }
71 inline ull get(int x,int l){return b[x]-b[x+l]*a[l];}
72
73 int main()
74
75 {
76
77 freopen("input.txt","r",stdin);
78
79 freopen("output.txt","w",stdout);
80
81 n=read();q=read();
82 scanf("%s",ch);s=ch;
83 a[0]=1;
84 for1(i,maxn-1)a[i]=a[i-1]*(ull)base;
85 rebuild();
86 while(q--)
87 {
88 scanf("%s",ch);
89 if(ch[0]==‘L‘)
90 {
91 int x=read()-1,y=read()-1,l=0,r=n-y;
92 while(l<=r)
93 {
94 int mid=(l+r)>>1;
95 if(get(x,mid)==get(y,mid))l=mid+1;else r=mid-1;
96 }
97 printf("%d\n",r);
98 }
99 else if(ch[0]==‘A‘)
100 {
101 int x=read()-1;
102 scanf("%s",ch);
103 s.insert(x,ch);
104 rebuild();
105 }
106 else if(ch[0]==‘C‘)
107 {
108 int x=read()-1,y=read()-1;
109 scanf("%s",ch);
110 for2(i,x,y)s[i]=ch[i-x];
111 rebuild();
112 }
113 else
114 {
115 int x=read()-1,y=read()-1;
116 s.erase(x,y-x+1);
117 rebuild();
118 }
119 }
120
121 return 0;
122
123 }
再贴一下splay的代码，sad story。。。
代码：
1 #include<cstdio>
2
3 #include<cstdlib>
4
5 #include<cmath>
6
7 #include<cstring>
8
9 #include<algorithm>
10
11 #include<iostream>
12
13 #include<vector>
14
15 #include<map>
16
17 #include<set>
18
19 #include<queue>
20
21 #include<string>
22
23 #define inf 1000000000
24
25 #define maxn 1000000+5
26
27 #define maxm 500+100
28
29 #define eps 1e-10
30
31 #define ull unsigned long long
32
33 #define pa pair<int,int>
34
35 #define for0(i,n) for(int i=0;i<=(n);i++)
36
37 #define for1(i,n) for(int i=1;i<=(n);i++)
38
39 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
40
41 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
42
43 #define mod 1000000007
44
45 using namespace std;
46
47 inline int read()
48
49 {
50
51 int x=0,f=1;char ch=getchar();
52
53 while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
54
55 while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}
56
57 return x*f;
58
59 }
60 int n,m,q,xx,yy,rt,tot,id[maxn],s[maxn],c[maxn][2],fa[maxn];
61 ull v[maxn],sum[maxn],hash[maxn];
62 char st[maxn];
63 inline void pushup(int x)
64 {
65 if(!x)return;
66 int l=c[x][0],r=c[x][1];
67 s[x]=s[l]+s[r]+1;
68 sum[x]=sum[r]+v[x]*hash[s[r]]+sum[l]*hash[s[r]+1];
69 //if(x==7)cout<<x<<‘ ‘<<l<<‘ ‘<<r<<‘ ‘<<sum[x]<<‘ ‘<<sum[r]<<‘ ‘<<s[r]<<‘ ‘<<v[x]<<endl;
70 }
71 inline void rotate(int x,int &k)
72 {
73 int y=fa[x],z=fa[y],l=c[y][1]==x,r=l^1;
74 if(y!=k)c[z][c[z][1]==y]=x;else k=x;
75 //cout<<x<<‘ ‘<<y<<‘ ‘<<z<<endl;
76 fa[x]=z;fa[y]=x;fa[c[x][r]]=y;
77 c[y][l]=c[x][r];c[x][r]=y;
78 pushup(y);pushup(x);
79 }
80 inline void splay(int x,int &k)
81 {
82 while(x!=k)
83 {
84 int y=fa[x],z=fa[y];
85 if(y!=k)
86 {
87 if((c[z][0]==y)^(c[y][0]==x))rotate(x,k);else rotate(y,k);
88 }
89 rotate(x,k);
90 }
91 }
92 void build(int l,int r,int f)
93 {
94 if(l>r)return;
95 int mid=(l+r)>>1,x=id[mid]=++tot,y=fa[x]=id[f];
96 v[x]=st[mid];c[y][mid>f]=x;
97 if(l==r)
98 {
99 sum[x]=v[x];s[x]=1;
100 return;
101 }
102 build(l,mid-1,mid);build(mid+1,r,mid);
103 pushup(x);
104 //cout<<x<<‘ ‘<<c[x][0]<<‘ ‘<<c[x][1]<<‘ ‘<<sum[x]<<endl;
105 }
106 inline int find(int x,int k)
107 {
108 int l=c[x][0],r=c[x][1];
109 if(s[l]+1==k)return x;
110 else if(s[l]>=k)return find(l,k);
111 else return find(r,k-s[l]-1);
112 }
113 inline void split(int l,int r)
114 {
115 xx=find(rt,l);yy=find(rt,r);
116 splay(xx,rt);splay(yy,c[xx][1]);
117 }
118 inline void print(int x)
119 {
120 if(!x)return;
121 //cout<<x<<‘ ‘<<c[x][0]<<‘ ‘<<c[x][1]<<"AAAAAAAAAAA"<<endl;
122 print(c[x][0]);
123 cout<<(char)v[x];
124 print(c[x][1]);
125 }
126 inline ull query(int l,int r)
127 {
128 split(l,r+2);
129 //cout<<l<<‘ ‘<<r<<endl;
130 //cout<<l<<‘ ‘<<r<<‘ ‘<<c[y][0]<<‘ ‘<<sum[c[y][0]]<<‘ ‘<<v[c[y][0]]<<endl;
131 //print(c[yy][0]);cout<<endl;
132 //cout<<sum[c[yy][0]]<<endl;
133 return sum[c[yy][0]];
134 }
135
136 int main()
137
138 {
139
140 n=read();q=read();
141 hash[0]=1;
142 for1(i,maxn-1)hash[i]=hash[i-1]*(ull)150;
143 scanf("%s",st+2);m=strlen(st+2);
144 st[1]=st[m+1+1]=‘a‘;
145 build(1,m+2,0);rt=id[(1+m+2)>>1];
146 //cout<<id[2]<<‘ ‘<<id[3]<<endl;
147 while(q--)
148 {
149 char ch[10];scanf("%s",ch);
150 //cout<<"AAAAAAAAAA"<<endl;
151 if(ch[0]==‘L‘)
152 {
153 int a=read(),b=read(),l=0,r=s[rt]-2-b+1;
154 while(l<=r)
155 {
156 int mid=(l+r)>>1;
157 //cout<<l<<‘ ‘<<mid<<‘ ‘<<r<<endl;
158 if(query(a,a+mid-1)==query(b,b+mid-1))l=mid+1;else r=mid-1;
159 }
160 printf("%d\n",r);
161 }
162 else if(ch[0]==‘A‘)
163 {
164 int a=read();//cout<<a<<endl;
165 scanf("%s",st+1);m=strlen(st+1);
166 build(1,m,0);
167 split(a,a+1);
168 fa[c[yy][0]=id[(1+m)>>1]]=yy;
169 pushup(yy);pushup(xx);
170 }
171 else if(ch[0]==‘C‘)
172 {
173 int a=read(),b=read();
174 scanf("%s",st+1);m=strlen(st+1);
175 build(1,m,0);
176 split(a,b+2);
177 fa[c[yy][0]=id[(1+m)>>1]]=yy;
178 pushup(yy);pushup(xx);
179 }
180 else
181 {
182 int a=read(),b=read();
183 split(a,b+2);
184 c[yy][0]=0;
185 pushup(yy);pushup(xx);
186 }
187 }
188
189 return 0;
190
191 }
时间： 2024-11-09 02:04:54