CodeForces985F -- Isomorphic Strings
3 seconds
256 megabytes
standard input
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:
- f(si) = ti for any index i,
- for any character
there is exactly one character
that f(x) = y, - 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.
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: xi, yi and leni (1 ≤ xi, yi ≤ n, 1 ≤ leni ≤ n - max(xi, yi) + 1) — the description of the pair of the substrings to check.
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.
7 4
abacaba
1 1 1
1 4 2
2 1 3
2 4 3
YES
YES
NO
YES
The queries in the example are following:
- substrings "a" and "a" are isomorphic: f(a) = a;
- substrings "ab" and "ca" are isomorphic: f(a) = c, f(b) = a;
- substrings "bac" and "aba" are not isomorphic since f(b) and f(c) must be equal to a at same time;
- substrings "bac" and "cab" are isomorphic: f(b) = c, f(a) = a, f(c) = b.
AC代码为:
#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;
}
CodeForces985F -- Isomorphic Strings的更多相关文章
- [LeetCode] Isomorphic Strings
Isomorphic Strings Total Accepted: 30898 Total Submissions: 120944 Difficulty: Easy Given two string ...
- leetcode:Isomorphic Strings
Isomorphic Strings Given two strings s and t, determine if they are isomorphic. Two strings are isom ...
- 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 ...
- [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 ...
- 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 ...
- 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... ...
- LeetCode 205. 同构字符串(Isomorphic Strings)
205. 同构字符串 205. Isomorphic Strings
- LeetCode_205. Isomorphic Strings
205. Isomorphic Strings Easy Given two strings s and t, determine if they are isomorphic. Two string ...
- 【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
- nyoj 92-图像有用区域 (BFS)
92-图像有用区域 内存限制:64MB 时间限制:3000ms 特判: No 通过数:4 提交数:12 难度:4 题目描述: “ACKing”同学以前做一个图像处理的项目时,遇到了一个问题,他需要摘取 ...
- 学习PHP框架只停留在会用层面,职业生涯肯定走不远!
工作这么多年,也面试过很多PHP工程师,我发现很多PHP工程师只停留在使用框架的层面,然而对框架底层根本没有深入去了解,那么这就会给自己的职业生涯带来一定的瓶颈,当遇到问题的时候你就无从下手,不知道如 ...
- 程序员用于机器学习编程的Python 数据处理库 pandas 进阶教程
数据访问 在入门教程中,我们已经使用过访问数据的方法.这里我们再集中看一下. 注:这里的数据访问方法既适用于Series,也适用于DataFrame. **基础方法:[]和. 这是两种最直观的方法,任 ...
- python基础-面向对象编程之封装、访问限制机制和property
面向对象编程之封装 封装 定义:将属性和方法一股脑的封装到对象中,使对象可通过"对象."的方式获取或存储数据. 作用:让对象有了"."的机制,存取数据更加方便 ...
- C语言与汇编语言混合编程实验
混合编程方法: 模块链接法 汇编指令嵌入法 1: 模块链接法则 模块链接法是指分别用汇编语言和C语言实现独立的模块(或子程序),再用链接程序把各模块生成的obj文件连接成一个可执行程序. 1:C语言调 ...
- linux hwclock
在Linux中,系统时间(软件时间)和硬件时间,并不会自动同步,系统时间和硬件时间以异步的方式运行,互不干扰.硬件时间的运行,是靠Bios电池来维持,而系统时间,是在系统开机的时候,会自动从Bios中 ...
- Rust入坑指南:鳞次栉比
很久没有挖Rust的坑啦,今天来挖一些排列整齐的坑.没错,就是要介绍一些集合类型的数据类型."鳞次栉比"这个标题是不是显得很有文化? 在Rust入坑指南:常规套路一文中我们已经介绍 ...
- Gitlab用户信息批量导出
前言 因运维体系中涉及到用户权限管理及统计,需将Gitlab用户数据提取出来并录入到公司内部自建的权限统计平台. 本文将对Gitlab的用户信息数据批量导出进行操作说明! 思路 A)要对数据进行批量的 ...
- vscode vue模版
{ "Print to console": { "prefix": "vue", "body": [ "< ...
- 【决战西二旗】|理解Sort算法
前言 前面两篇文章介绍了快速排序的基础知识和优化方向,今天来看一下STL中的sort算法的底层实现和代码技巧. 众所周知STL是借助于模板化来支撑数据结构和算法的通用化,通用化对于C++使用者来说已经 ...