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。。。  字符串哈希， 将26个字符分开来hash， 那么check就变成啦， 区间内对应的26个字符的hash值是否一致。

即如果 a -> b  那么区间1内a的hash值等于区间2内b的hash值。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int,int>
#define piii pair<int, pair<int,int>>

using namespace std;

const int N = 2e5 + 7;
const int M = 1e4 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;

int n, m;
LL hs[26][N], f[N], base = 17, seed = 11;
char s[N];

vector<int> vec[26];
LL getHash(int x, int l, int r) {
    LL ans = hs[x][l];
    if(r < n - 1) ans -= hs[x][r + 1] * f[r - l + 1];
    ans = (ans % mod + mod) % mod;
    return ans;
}

bool check(int l1, int r1, int l2, int r2) {
    for(int j = 0; j < 26; j++) {
        int p = lower_bound(vec[j].begin(), vec[j].end(), l1) - vec[j].begin();
        if(p >= vec[j].size() || vec[j][p] > r1) continue;
        if(getHash(j, l1, r1) != getHash(s[l2 + (vec[j][p] - l1)] - 'a', l2, r2))
            return false;
    }
    return true;
}
int main() {

    f[0] = 1; for(int i = 1; i < N; i++) f[i] = f[i - 1] * base % mod;
    scanf("%d%d", &n, &m);
    scanf("%s", s);
    n = strlen(s);


    for(int i = n - 1; i >= 0; i--) {
        for(int j = 0; j < 26; j++) {
            hs[j][i] = hs[j][i + 1] * base % mod;
            if(j == s[i] - 'a') {
                hs[j][i] = (hs[j][i] + seed) % mod;
            }
        }
    }

    for(int i = 0; i < n; i++)
        vec[s[i] - 'a'].push_back(i);

    while(m--) {
        int x, y, len;
        scanf("%d%d%d", &x, &y, &len);
        int l1 = x - 1, l2 = y - 1, r1 = l1 + len - 1, r2 = l2 + len - 1;

        if(check(l1, r1, l2, r2)) puts("YES");
        else puts("NO");
    }
    return 0;
}

/*
*/