CoderForces985F-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.
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
- [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... ...
- CodeForces985F -- Isomorphic Strings
F. Isomorphic Strings time limit per test 3 seconds memory limit per test 256 megabytes input standa ...
- 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
- python——切片
切片 格式:[start : end : step] start:起始索引,从0开始,-1表示结束 end:结束索引 step:步长 # 字符串,列表,元组等都可以支持切片截取的操作# 切片必须依赖于 ...
- Download-学习资源下载
小白求关爱:链接为本人工作中学习所碰到的,如有不对之处,请及时联系加以改正,后续仍会追加新链接地址并及时更新旧链接地址 Jdk历史版本 https://www.oracle.com/technetwo ...
- 这份最新Python面试精选问题你会几道?
相信很多小伙伴学python以后都想进大厂,但是进大厂前你得了解些大厂面试题,可以在面试前复习下,以下是精选的5道python面试题: 第一. Python 的特点和优点是什么? Python 可以作 ...
- 100天搞定机器学习|Day56 随机森林工作原理及调参实战(信用卡欺诈预测)
本文是对100天搞定机器学习|Day33-34 随机森林的补充 前文对随机森林的概念.工作原理.使用方法做了简单介绍,并提供了分类和回归的实例. 本期我们重点讲一下: 1.集成学习.Bagging和随 ...
- C. Present(二分 + 扫描线)
题目链接:http://codeforces.com/contest/460/problem/C 题意: n盆花,浇k次水, 每次可使花高度 + 1, 每次可浇相邻的w盆,ai 表示 i-th盆花 的 ...
- mysql基础之约束
约束的目的: 1.约束保证数据的完整性和一致性. 2.约束分为表级约束 和 列级 约束.(针对约束字段的数目的多少来确定的) 3.约束类型包括 not null (非空约束) primary key( ...
- 彻底搞懂CSS伪类选择器:is、not
本文介绍一下Css伪类:is和:not,并解释一下is.not.matches.any之前的关系 :not The :not() CSS pseudo-class represents element ...
- eNSP仿真软件之配置Trunk口
实验原理 在以太网中,通过划分VLAN来隔离广播域和增强网络通信的安全性.以太网通常由多台交换机组成,为了使VLAN的数据帧跨越多台交换机传递,交换机之间互连的链路需要配置为干道链路(Trunk Li ...
- 2019-9-25:渗透测试,基础学习,medusa爆破学习
Medusa(美杜莎)暴力破解使用 该文章仅供学习,利用方法来自网络文章,仅供参考 一.简介 medusa(美杜莎)是一个速度快,支持大规模并行,模块化,爆破登陆,可以同时对多个主机,用户或是密码执行 ...
- 学习Python第一天:找了4本专属小白的书籍(前期入门打基础)
我们提供一个初学者最好的Python书籍列表.Python是一个初级程序员可以学习编程的最友好语言之一.为了帮助您开始使用Python编程,我们分享此列表.泡一杯茶,选一本书阅读,开始使用Python ...