高级哈希
掌握高级哈希技术,包括字符串哈希、滚动哈希和多项式哈希,用于竞赛编程中的高效字符串匹配和比较。
1. 字符串哈希
多项式滚动哈希用于高效的字符串比较和子串匹配,经过 O(n) 预处理后可实现 O(1) 复杂度。
滚动哈希实现
#include <iostream>
#include <vector>
#include <string>
using namespace std;
class RollingHash {
private:
static const long long MOD = 1e9 + 7;
static const long long BASE = 31;
vector<long long> hash, power;
public:
RollingHash(const string& s) {
int n = s.length();
hash.resize(n + 1);
power.resize(n + 1);
power[0] = 1;
for (int i = 1; i <= n; i++) {
power[i] = (power[i-1] * BASE) % MOD;
}
hash[0] = 0;
for (int i = 1; i <= n; i++) {
hash[i] = (hash[i-1] * BASE + (s[i-1] - 'a' + 1)) % MOD;
}
}
long long getHash(int l, int r) { // 0-indexed, inclusive
l++; r++; // Convert to 1-indexed
long long result = hash[r] - (hash[l-1] * power[r-l+1]) % MOD;
return (result % MOD + MOD) % MOD;
}
bool areEqual(int l1, int r1, int l2, int r2) {
return getHash(l1, r1) == getHash(l2, r2);
}
};
int main() {
string s = "abacaba";
RollingHash rh(s);
cout << "String: " << s << endl;
// Check if substrings are equal
cout << "s[0:2] == s[4:6]: " << rh.areEqual(0, 2, 4, 6) << endl; // "aba" == "aba"
cout << "s[0:1] == s[3:4]: " << rh.areEqual(0, 1, 3, 4) << endl; // "ab" == "ca"
cout << "s[1:3] == s[5:6]: " << rh.areEqual(1, 3, 5, 6) << endl; // "bac" == "ba"
// Get hash values
cout << "Hash of s[0:2]: " << rh.getHash(0, 2) << endl;
cout << "Hash of s[4:6]: " << rh.getHash(4, 6) << endl;
return 0;
}2. 双重哈希
使用两个不同的哈希函数来减少冲突概率并提高字符串比较的可靠性。
双重哈希实现
#include <iostream>
#include <vector>
#include <string>
using namespace std;
class DoubleHash {
private:
static const long long MOD1 = 1e9 + 7;
static const long long MOD2 = 1e9 + 9;
static const long long BASE1 = 31;
static const long long BASE2 = 37;
vector<long long> hash1, hash2, power1, power2;
public:
DoubleHash(const string& s) {
int n = s.length();
hash1.resize(n + 1);
hash2.resize(n + 1);
power1.resize(n + 1);
power2.resize(n + 1);
power1[0] = power2[0] = 1;
for (int i = 1; i <= n; i++) {
power1[i] = (power1[i-1] * BASE1) % MOD1;
power2[i] = (power2[i-1] * BASE2) % MOD2;
}
hash1[0] = hash2[0] = 0;
for (int i = 1; i <= n; i++) {
int c = s[i-1] - 'a' + 1;
hash1[i] = (hash1[i-1] * BASE1 + c) % MOD1;
hash2[i] = (hash2[i-1] * BASE2 + c) % MOD2;
}
}
pair<long long, long long> getHash(int l, int r) {
l++; r++;
long long h1 = hash1[r] - (hash1[l-1] * power1[r-l+1]) % MOD1;
long long h2 = hash2[r] - (hash2[l-1] * power2[r-l+1]) % MOD2;
h1 = (h1 % MOD1 + MOD1) % MOD1;
h2 = (h2 % MOD2 + MOD2) % MOD2;
return {h1, h2};
}
bool areEqual(int l1, int r1, int l2, int r2) {
return getHash(l1, r1) == getHash(l2, r2);
}
};
int main() {
string s = "programming";
DoubleHash dh(s);
cout << "String: " << s << endl;
auto h1 = dh.getHash(0, 3); // "prog"
auto h2 = dh.getHash(7, 10); // "ming"
cout << "Hash of 'prog': (" << h1.first << ", " << h1.second << ")" << endl;
cout << "Hash of 'ming': (" << h2.first << ", " << h2.second << ")" << endl;
cout << "Are 'prog' and 'ming' equal: " << dh.areEqual(0, 3, 7, 10) << endl;
cout << "Are 'prog' and 'prog' equal: " << dh.areEqual(0, 3, 0, 3) << endl;
return 0;
}总结
- 滚动哈希:经过 O(n) 预处理后实现 O(1) 子串比较
- 双重哈希:使用两个哈希函数减少冲突概率
- 应用:字符串匹配、重复检测、模式搜索
- 复杂度:O(n) 空间和预处理,O(1) 查询时间