Description
You have a string and queries of two types:
replace i’th character of the string by character a;
check if substring sj...sk is a palindrome.
Input
The first line contains a string consisting of n small English letters. The second line contains an integer m that is the number of queries (5 ≤ n, m ≤ 105). The next m lines contain the queries.
Each query has either form “change i a”, or “palindrome? j k”, where i, j, k are integers (1 ≤ i ≤ n; 1 ≤ j ≤ k ≤ n), and character a is a small English letter.
Output
To all second type queries, you should output “Yes” on a single line if substring sj...sk is a palindrome and “No” otherwise.
Sample input
abcda
5
palindrome? 1 5
palindrome? 1 1
change 4 b
palindrome? 1 5
palindrome? 2 4
Sample onput
No
Yes
Yes
Yes
题解
使用线段树维护哈希值。
滚动哈希即rabin-karp
原hash:
(ha[i]=ha[i-1] imes seed+s[i])
可以预处理出seed进位:
(p[i]=p[i-1] imes seed)
这样每个位置的哈希值可以直接算出。
如果要改变某个位置的字符，那么原hash值变成
(ha[i]=p[i-1] imes ch)
这样的hash计算无需滚动，因为计算出了进位，每一位是直接算出的。那么可以直接合并两个区间的哈希值，
即：
ha[rt]=ha[rt<<1]+ha[rt<<1|1]
改变也是单调修改，这些都是线段树的经典应用
import java.io.*;
import java.util.*;
public class Main {
static final int N = (int)1e5+10;
//static final long MOD=(int)1e18+7;
static final int inf = 0x3f3f3f3f;
static char a[]=new char[N];
static long p[]=new long[N];
static long ha[]=new long[N];
static long sum[][]=new long[2][N<<2];
static void Init(int n) {
int seed=123;
p[0]=1;
for(int i=1;i<=n;i++) p[i]=(p[i-1]*seed);
}
static void PushUp(int rt,int flag) {
sum[flag][rt]=(sum[flag][rt<<1]+sum[flag][rt<<1|1]);
}
static void Build(int rt,int l,int r,int flag) {
if(l==r) {
sum[flag][rt]=ha[l];
return;
}
int mid=(l+r)>>1;
Build(rt<<1,l,mid,flag);
Build(rt<<1|1,mid+1,r,flag);
PushUp(rt,flag);
}
static void update(int a,long b,int rt,int l,int r,int flag) {
if(l==r) {
sum[flag][rt]=b;
return;
}
int mid=(l+r)>>1;
if(a<=mid) update(a,b,rt<<1,l,mid,flag);
if(a>mid) update(a,b,rt<<1|1,mid+1,r,flag);
PushUp(rt,flag);
}
static long query(int a,int b,int rt,int l,int r,int flag) {
if(a<=l&&r<=b) {
return sum[flag][rt];
}
int mid=(l+r)>>1;
long res=0;
if(a<=mid) res=(res+query(a,b,rt<<1,l,mid,flag));
if(b>mid) res=(res+query(a,b,rt<<1|1,mid+1,r,flag));
return res;
}
public static void main(String[] args) {
InputStream sysi = System.in;
OutputStream syso = System.out;
InputReader in = new InputReader(sysi);
PrintWriter out = new PrintWriter(syso);
a=in.next().toCharArray();
int m=in.nextInt();
int s,e,s1,e1,n=a.length;char ch;
Init(n);
for(int i=0;i<n;i++) ha[i+1]=((long)a[i]*p[i]);
Build(1,1,n,0);
char tmp;
for(int i=0;i<n/2;i++) {
tmp=a[i];a[i]=a[n-i-1];
a[n-i-1]=tmp;
}
for(int i=0;i<n;i++) ha[i+1]=((long)a[i]*p[i]);
Build(1,1,n,1);String str;
while(m--!=0) {
if(in.next().equals("palindrome?")) {
s=in.nextInt();e=in.nextInt();
long ha1=query(s,e,1,1,n,0);
s1=n+1-e;e1=n+1-s;
long ha2=query(s1,e1,1,1,n,1);
if(s1>s) ha1=(ha1*p[s1-s]);
else ha2=(ha2*p[s-s1]);
if(ha1==ha2) out.println("Yes");
else out.println("No");
}else {
s=in.nextInt();str=in.next();
ch=str.charAt(0);
update(s,p[s-1]*ch,1,1,n,0);
update(n+1-s,p[n-s]*ch,1,1,n,1);
}
out.flush();
}
out.close();
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
}