CodeForces985F -- Isomorphic Strings
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;
}