Codeforces 985 F

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

代码：

#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pii pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = 2e5 + 5;
const int MOD = 1e9 + 7;
const int base = 13;
char s[N];
LL Hash[N][26], p[N], t[30], tt[30];
LL get_H(int l, int r, int i) {
    return ((Hash[r][i] - Hash[l-1][i]*p[r-l+1])%MOD + MOD)%MOD;
}
bool check(int x, int y, int len) {
    for (int i = 0; i < 26; i++) {
        t[i] = get_H(x, x+len-1, i);
        tt[i] = get_H(y, y+len-1, i);
    }
    sort(t, t+26);
    sort(tt, tt+26);
    for (int i = 0; i < 26; i++) if(t[i] != tt[i]) return false;
    return true;
}
int main() {
    int n, m, x, y, len;
    scanf("%d%d", &n, &m);
    scanf("%s", s+1);
    p[0] = 1;
    for (int i = 1; i <= n; i++) p[i] = (p[i-1] * base) % MOD;
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j < 26; j++) {
            Hash[i][j] = (Hash[i-1][j] * base + (s[i] == 'a'+j))%MOD;
        }
    }
    while(m--) {
        scanf("%d%d%d", &x, &y, &len);
        if(check(x, y, len)) puts("YES");
        else puts("NO");
    }
    return 0;
}