CF985F Isomorphic Strings

题目描述

You are given a string $s$ of length $n$ consisting of lowercase English letters.

For two given strings $s$ and $t$ , say $S$ is the set of distinct characters of $s$ and $T$ is the set of distinct characters of $t$ . The strings $s$ and $t$ are isomorphic if their lengths are equal and there is a one-to-one mapping (bijection) $f$ between $S$ and $T$ for which $f(s^i)=t^i$ . Formally:

1. $f(s^i)=t^i$ for any index $i$ ,
2. for any character there is exactly one character that $f(x)=y$ ,
3. for any character there is exactly one character that $f(x)=y$ .

For example, the strings "aababc" and "bbcbcz" are isomorphic. Also the strings "aaaww" and "wwwaa" are isomorphic. The following pairs of strings are not isomorphic: "aab" and "bbb", "test" and "best".

You have to handle $m$ queries characterized by three integers $x,y,len$ ( $1<=x,y<=n-len+1$ ). For each query check if two substrings $s[x...x+len-1]$ and $s[y...y+len-1]$ are isomorphic.

输入输出格式

输入格式：

The first line contains two space-separated integers $n$ and $m$ ( $1<=n<=2\cdot 10^5$ , $1<=m<=2\cdot 10^5$ ) — the length of the string $s$ and the number of queries.

The second line contains string $s$ consisting of $n$ lowercase English letters.

The following $m$ lines contain a single query on each line: $x^i$ , $y^i$ and $le{n}^i$ ( $1<=x^i,y^i<=n$ , $1<=le{n}^i<=n-max(x^i,y^i)+1$ ) — the description of the pair of the substrings to check.

输出格式：

For each query in a separate line print "YES" if substrings $s[x^i...x^i+le{n}^i-1]$ and $s[y^i...y^i+le{n}^i-1]$ are isomorphic and "NO" otherwise.

输入输出样例

输入样例#1： 

7 4
abacaba
1 1 1
1 4 2
2 1 3
2 4 3

输出样例#1： 

YES
YES
NO
YES

说明

The queries in the example are following:

1. substrings "a" and "a" are isomorphic: $f(a)=a$ ;
2. substrings "ab" and "ca" are isomorphic: $f(a)=c$ , $f(b)=a$ ;
3. substrings "bac" and "aba" are not isomorphic since $f(b)$ and $f(c)$ must be equal to $a$ at same time;
4. substrings "bac" and "cab" are isomorphic: $f(b)=c$ , $f(a)=a$ , $f(c)=b$ .

 

Solution：

　　本题还是HRZ学长讲的算法，也是贼有意思。

　　题意中的相似就并不在意字符具体是什么，而在于字符的排列情况是否能一致，而若我们用$26$个字母分别为关键字弄出$26$个$0,1$串，那么一个字符串的相对位置关系是可以用这$26$个$0,1$排列来唯一确定。

　　于是，我们对整个字符串分别以$26$个字母为关键字进行hash，解释一下关键字意思是比如$a$字符为关键字，那么所有非$a$的字符所对应的$K$进制位为$0$，只有为该关键字的位为$1$，这样就能保证相似的排列对应的hash值一致，而整个hash过程也就$O(n)$递推一遍就好了。

　　然后对于每次询问，我们就将两段字符串的$26$个hash值取出来，分别用$a,b$数组存下来，从小到大排序，然后比较是否相等即可。

　　注意本题hack模数$998244353$，所以我改用了$19260817$做单模数，当然更建议用多模数减少错误率。

代码：

#include<bits/stdc++.h>
#define il inline
#define ll long long
#define For(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)
#define Bor(i,a,b) for(int (i)=(b);(i)>=(a);(i)--)
using namespace std;
const int N=200005,M=131,mod2=998244353,mod1=19260817;
int n,m,x,y,len;
ll f[N][27],sum[N],a[27],b[27];
char s[N];

il int gi(){
    int a=0;char x=getchar();bool f=0;
    while((x<'0'||x>'9')&&x!='-')x=getchar();
    if(x=='-')x=getchar(),f=1;
    while(x>='0'&&x<='9')a=(a<<3)+(a<<1)+x-48,x=getchar();
    return f?-a:a;
}

il bool check(){
    For(i,1,26) 
        a[i]=(f[x+len-1][i]-f[x-1][i]*sum[len]%mod1+mod1)%mod1,
        b[i]=(f[y+len-1][i]-f[y-1][i]*sum[len]%mod1+mod1)%mod1;
    sort(a+1,a+27),sort(b+1,b+27);
    For(i,1,26) if(a[i]!=b[i])return 0;
    return 1;
}

int main(){
    n=gi(),m=gi();
    scanf("%s",s+1);
    sum[0]=1;
    For(i,1,n) {
        sum[i]=(sum[i-1]*M)%mod1;
        For(j,1,26) f[i][j]=(f[i-1][j]*M%mod1+(s[i]=='a'+j-1?1:0))%mod1;    
    }
    while(m--){
        x=gi(),y=gi(),len=gi();
        (check())?puts("YES"):puts("NO");
    }
    return 0;
}