F. Isomorphic Strings
time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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(si) = ti. Formally:

  1. f(si) = ti 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.

Input

The first line contains two space-separated integers n and m (1 ≤ n ≤ 2·105, 1 ≤ m ≤ 2·105) — 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: xiyi and leni (1 ≤ xi, yi ≤ n, 1 ≤ leni ≤ n - max(xi, yi) + 1) — the description of the pair of the substrings to check.

Output

For each query in a separate line print "YES" if substrings s[xi... xi + leni - 1] and s[yi... yi + leni - 1] are isomorphic and "NO" otherwise.

Example
input
Copy
7 4
abacaba
1 1 1
1 4 2
2 1 3
2 4 3
output
Copy
YES
YES
NO
YES
Note

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) = cf(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) = cf(a) = af(c) = b.

题意:在一个字符串里随意找两个长度相同的子串,比较是否为同构字符串;

字符串hash,比较两个字符串的每个字符的hash值是否相同;

AC;

#include<iostream>
#include<queue>
#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn=2e5+10;
const ll MOD=1e9+7;
ll h[26][maxn],x=107,px[maxn];
char s[maxn];
int main()
{
int n,m;
scanf("%d%d", &n,&m);
scanf("%s", s+1);
px[0]=1;
for(int i = 1; i <= n; ++i) px[i]=px[i-1]*x%MOD;
for(int i = 0; i < 26; ++i)
{
for(int j = 1; j <= n; ++j) h[i][j]=(h[i][j-1]*x+int(s[j] == (i+'a')))%MOD;
}
while(m--)
{
int x,y,l;
scanf("%d%d%d", &x,&y,&l);
vector<int> p,q;
for(int i = 0; i < 26; ++i)
{
p.push_back(((h[i][x+l-1]-px[l]*h[i][x-1]%MOD)%MOD+MOD)%MOD);
q.push_back(((h[i][y+l-1]-px[l]*h[i][y-1]%MOD)%MOD+MOD)%MOD);
}
sort(q.begin(),q.end());sort(p.begin(),p.end());
printf("%s\n", p==q? "YES":"NO");
}
return 0;
}

字符串

CoderForces985F-Isomorphic Strings的更多相关文章

  1. [LeetCode] Isomorphic Strings

    Isomorphic Strings Total Accepted: 30898 Total Submissions: 120944 Difficulty: Easy Given two string ...

  2. leetcode:Isomorphic Strings

    Isomorphic Strings Given two strings s and t, determine if they are isomorphic. Two strings are isom ...

  3. Codeforces Educational Codeforces Round 44 (Rated for Div. 2) F. Isomorphic Strings

    Codeforces Educational Codeforces Round 44 (Rated for Div. 2) F. Isomorphic Strings 题目连接: http://cod ...

  4. [leetcode]205. Isomorphic Strings 同构字符串

    Given two strings s and t, determine if they are isomorphic. Two strings are isomorphic if the chara ...

  5. Codeforces 985 F - Isomorphic Strings

    F - Isomorphic Strings 思路:字符串hash 对于每一个字母单独hash 对于一段区间,求出每个字母的hash值,然后排序,如果能匹配上,就说明在这段区间存在字母间的一一映射 代 ...

  6. Educational Codeforces Round 44 (Rated for Div. 2) F - Isomorphic Strings

    F - Isomorphic Strings 题目大意:给你一个长度为n 由小写字母组成的字符串,有m个询问, 每个询问给你两个区间, 问你xi,yi能不能形成映射关系. 思路:这个题意好难懂啊... ...

  7. CodeForces985F -- Isomorphic Strings

    F. Isomorphic Strings time limit per test 3 seconds memory limit per test 256 megabytes input standa ...

  8. LeetCode 205. 同构字符串(Isomorphic Strings)

    205. 同构字符串 205. Isomorphic Strings

  9. LeetCode_205. Isomorphic Strings

    205. Isomorphic Strings Easy Given two strings s and t, determine if they are isomorphic. Two string ...

  10. 【刷题-LeetCode】205. Isomorphic Strings

    Isomorphic Strings Given two strings *s* and *t*, determine if they are isomorphic. Two strings are ...

随机推荐

  1. show语句大全

    基于本人对MySQL的使用,现将常用的MySQL show 语句列举如下: 1.show databases ; // 显示mysql中所有数据库的名称 2.show tables [from dat ...

  2. Proxy动态代理-增强方法

    增强对象的功能 设计模式:一些通用的解决固定问题的方式 装饰器模式 代理模式 概念: 在代理模式(Proxy Pattern)中,一个类代表另一个类的功能.这种类型的设计模式属于结构型模式. 在代理模 ...

  3. 淘宝小练习#css

    * { margin: 0; padding: 0; } a { text-decoration: none; } .box { background: #f4f4f4; } /* 头部样式STAR ...

  4. 一张图讲解最少机器搭建FastDFS高可用分布式集群安装说明

     很幸运参与零售云快消平台的公有云搭建及孵化项目.零售云快消平台源于零售云家电3C平台私有项目,是与公司业务强耦合的.为了适用于全场景全品类平台,集团要求项目平台化,我们抢先并承担了此任务.并由我来主 ...

  5. [剑指offer] 二叉搜索树的后序遍历序列 (由1个后续遍历的数组判断它是不是BST)

    ①题目 输入一个整数数组,判断该数组是不是某二叉搜索树的后序遍历的结果.如果是则输出Yes,否则输出No.假设输入的数组的任意两个数字都互不相同. ②思路 1.后续遍历的数组里,最后一个元素是根. 2 ...

  6. Spring-boot构建多模块依赖工程时,maven打包异常:程序包xxx不存在

    在qizhi项目改版的时候, 所有代码都迁移好了, 但是compile的时候报程序包*****不存在, 具体到某一个类就是: 找不到符号. 下面这篇文章是正解 http://hbxflihua.ite ...

  7. Mssql 查询某记录前后N条

    Sqlserver 查询指定记录前后N条,包括当前数据 条件 [ID] 查询 [N]条 select * from [Table] where ID in (select top ([N]+1) ID ...

  8. hdu 1205 吃糖果 (抽屉原理<鸽笼原理>)

    吃糖果Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submissi ...

  9. 深度剖析Javascript执行环境、作用域链

    一.执行环境 执行环境(也叫做执行上下文,Execution Context)是Javascript中最为重要的一个概念.执行环境定义了变量或函数有权访问其他数据,决定了它们各自的行为.每个执行环境都 ...

  10. 在lldb调试中调用c++函数 - 如何使用QuartzCore里面的日志消息

    承接上一篇,上一篇讲到可以在lldb调试中调用QuartzCore.framework里的CA::Render::Object::show方法来是观察CA::Render模块内的类的信息,但是在lld ...