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。

 1 #include<bits/stdc++.h>
 2 #define LL long long
 3 #define fi first
 4 #define se second
 5 #define mk make_pair
 6 #define pii pair<int,int>
 7 #define piii pair<int, pair<int,int>>
 8 
 9 using namespace std;
10 
11 const int N = 2e5 + 7;
12 const int M = 1e4 + 7;
13 const int inf = 0x3f3f3f3f;
14 const LL INF = 0x3f3f3f3f3f3f3f3f;
15 const int mod = 1e9 + 7;
16 
17 int n, m;
18 LL hs[26][N], f[N], base = 17, seed = 11;
19 char s[N];
20 
21 vector<int> vec[26];
22 LL getHash(int x, int l, int r) {
23     LL ans = hs[x][l];
24     if(r < n - 1) ans -= hs[x][r + 1] * f[r - l + 1];
25     ans = (ans % mod + mod) % mod;
26     return ans;
27 }
28 
29 bool check(int l1, int r1, int l2, int r2) {
30     for(int j = 0; j < 26; j++) {
31         int p = lower_bound(vec[j].begin(), vec[j].end(), l1) - vec[j].begin();
32         if(p >= vec[j].size() || vec[j][p] > r1) continue;
33         if(getHash(j, l1, r1) != getHash(s[l2 + (vec[j][p] - l1)] - 'a', l2, r2))
34             return false;
35     }
36     return true;
37 }
38 int main() {
39 
40     f[0] = 1; for(int i = 1; i < N; i++) f[i] = f[i - 1] * base % mod;
41     scanf("%d%d", &n, &m);
42     scanf("%s", s);
43     n = strlen(s);
44 
45 
46     for(int i = n - 1; i >= 0; i--) {
47         for(int j = 0; j < 26; j++) {
48             hs[j][i] = hs[j][i + 1] * base % mod;
49             if(j == s[i] - 'a') {
50                 hs[j][i] = (hs[j][i] + seed) % mod;
51             }
52         }
53     }
54 
55     for(int i = 0; i < n; i++)
56         vec[s[i] - 'a'].push_back(i);
57 
58     while(m--) {
59         int x, y, len;
60         scanf("%d%d%d", &x, &y, &len);
61         int l1 = x - 1, l2 = y - 1, r1 = l1 + len - 1, r2 = l2 + len - 1;
62 
63         if(check(l1, r1, l2, r2)) puts("YES");
64         else puts("NO");
65     }
66     return 0;
67 }
68 
69 /*
70 */