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

题目连接：

http://codeforces.com/contest/985/problem/F

Description

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:

f(s_i) = t_i for any index i,
for any character $\text{[math]}$ there is exactly one character $\text{[math]}$ that f(x) = y,
for any character $\text{[math]}$ there is exactly one character $\text{[math]}$ 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.

Sample Input

Sample Output

YES
YES
NO
YES

题意

判断字符串是否同构

Judging two string have the same structure or not.

题解：

将字符串抽象成26个01串，如果对应字母的26个对应位置的子串相同，则字符串同构。

Create 26 01-string, if the substring which is locate at the same position, two string have the same structure.

代码

#include <bits/stdc++.h>

using namespace std;

int n, k;
string d;
bool a[50][200010];
vector<int> v[26];
long long hs[50][200010];
long long c[200010];

const long long inf = 0x3f3f3f3f;
const long long MOD = 1e9 + 7;


int main() {
    cin >> n >> k >> d;
    d = " " + d;
    c[0] = 1;
    for (int i = 1; i < d.size(); i++) {
        a[d[i] - 'a'][i] = 1;
        v[d[i] - 'a'].push_back(i);
        c[i] = (c[i - 1] * inf) % MOD;
        for (int j = 0; j < 26; j++)
            hs[j][i] = (hs[j][i - 1] * inf + a[j][i]) % MOD;
    }

    for (int i = 1; i <= k; i++) {
        int x, y, z;
        cin >> x >> y >> z;
        bool flag = 1;
        for (int j = 0; j < 26; j++) {
            int pos = lower_bound(v[j].begin(), v[j].end(), x )-v[j].begin();
            if (pos<v[j].size())
                pos = v[j][pos];
            else
                continue;
            if (pos >= x + z) continue;
            int dual = d[pos + y - x]-'a';
            long long h1 = ((hs[j][x + z - 1] - hs[j][x - 1] * c[z]) % MOD + MOD) % MOD;
            long long h2 = ((hs[dual][y + z - 1] - hs[dual][y - 1] * c[z]) % MOD + MOD) % MOD;
            if (h1 != h2) {
                flag = 0;
                break;
            }
        }
        if (flag)
            cout << "YES" << endl;
        else
            cout << "NO" << endl;
    }

}