【URAL 1989】 Subpalindromes（线段树维护哈希）

Description

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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

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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

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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:

\(ha[i]=ha[i-1]\times seed+s[i]\)

可以预处理出seed进位:

\(p[i]=p[i-1]\times seed\)

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

\(ha[i]=p[i-1]\times ch\)

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());
		}
	}
}