CF1200E Compress Words 题解
题目链接:CF 或者 洛谷
注意到总字符串长度不超过 \(1e6\),对于两个串之间找前后缀匹配,只要能暴力枚举长度,\(check\ 为\ O(1)\),那么最后显然线性复杂度。可以考虑 \(kmp\),也可以考虑字符串哈希,最好上双哈希,然后拼串显然是在尾部继续添加新的前缀哈希,这个需要添加的串可以单独由匹配长度拿出来。匹配长度显然由最长开始枚举,直到答案串尾部的双哈希与待匹配串的前缀双哈希相同为止,空串特判。
参照代码
#include <bits/stdc++.h>
// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
// #define isPbdsFile
#ifdef isPbdsFile
#include <bits/extc++.h>
#else
#include <ext/pb_ds/priority_queue.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/trie_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/list_update_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/exception.hpp>
#include <ext/rope>
#endif
using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> tii;
typedef tuple<ll, ll, ll> tll;
typedef unsigned int ui;
typedef unsigned long long ull;
#define hash1 unordered_map
#define hash2 gp_hash_table
#define hash3 cc_hash_table
#define stdHeap std::priority_queue
#define pbdsHeap __gnu_pbds::priority_queue
#define sortArr(a, n) sort(a+1,a+n+1)
#define all(v) v.begin(),v.end()
#define yes cout<<"YES"
#define no cout<<"NO"
#define Spider ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define MyFile freopen("..\\input.txt", "r", stdin),freopen("..\\output.txt", "w", stdout);
#define forn(i, a, b) for(int i = a; i <= b; i++)
#define forv(i, a, b) for(int i=a;i>=b;i--)
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
#define endl '\n'
//用于Miller-Rabin
[[maybe_unused]] static int Prime_Number[13] = {0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};
template <typename T>
int disc(T* a, int n)
{
return unique(a + 1, a + n + 1) - (a + 1);
}
template <typename T>
T lowBit(T x)
{
return x & -x;
}
template <typename T>
T Rand(T l, T r)
{
static mt19937 Rand(time(nullptr));
uniform_int_distribution<T> dis(l, r);
return dis(Rand);
}
template <typename T1, typename T2>
T1 modt(T1 a, T2 b)
{
return (a % b + b) % b;
}
template <typename T1, typename T2, typename T3>
T1 qPow(T1 a, T2 b, T3 c)
{
a %= c;
T1 ans = 1;
for (; b; b >>= 1, (a *= a) %= c) if (b & 1) (ans *= a) %= c;
return modt(ans, c);
}
template <typename T>
void read(T& x)
{
x = 0;
T sign = 1;
char ch = getchar();
while (!isdigit(ch))
{
if (ch == '-') sign = -1;
ch = getchar();
}
while (isdigit(ch))
{
x = (x << 3) + (x << 1) + (ch ^ 48);
ch = getchar();
}
x *= sign;
}
template <typename T, typename... U>
void read(T& x, U&... y)
{
read(x);
read(y...);
}
template <typename T>
void write(T x)
{
if (typeid(x) == typeid(char)) return;
if (x < 0) x = -x, putchar('-');
if (x > 9) write(x / 10);
putchar(x % 10 ^ 48);
}
template <typename C, typename T, typename... U>
void write(C c, T x, U... y)
{
write(x), putchar(c);
write(c, y...);
}
template <typename T11, typename T22, typename T33>
struct T3
{
T11 one;
T22 tow;
T33 three;
bool operator<(const T3 other) const
{
if (one == other.one)
{
if (tow == other.tow) return three < other.three;
return tow < other.tow;
}
return one < other.one;
}
T3()
{
one = tow = three = 0;
}
T3(T11 one, T22 tow, T33 three) : one(one), tow(tow), three(three)
{
}
};
template <typename T1, typename T2>
void uMax(T1& x, T2 y)
{
if (x < y) x = y;
}
template <typename T1, typename T2>
void uMin(T1& x, T2 y)
{
if (x > y) x = y;
}
constexpr int N = 1e6 + 10;
constexpr int MOD[2] = {static_cast<int>(1e9) + 7, static_cast<int>(1e9) + 9};
constexpr ll Base[2] = {31, 33};
int basePow[N][2];
int pre[N][2];
int len;
string ans;
inline void add(const string& s)
{
ans += s;
for (const auto c : s)
{
forn(j, 0, 1) pre[len + 1][j] = (pre[len][j] * Base[j] + c) % MOD[j];
++len;
}
}
inline pii sub(const int l, const int r)
{
if (l > r) return pii(0, 0);
int x = modt(pre[r][0] - 1LL * pre[l - 1][0] * basePow[r - l + 1][0] % MOD[0], MOD[0]);
int y = modt(pre[r][1] - 1LL * pre[l - 1][1] * basePow[r - l + 1][1] % MOD[1], MOD[1]);
return pii(x, y);
}
inline void check(const string& s)
{
const int n = s.length();
vector tmpPre(n + 1, pii(0, 0));
forn(i, 1, n)
{
tmpPre[i].first = (tmpPre[i - 1].first * Base[0] + s[i - 1]) % MOD[0];
tmpPre[i].second = (tmpPre[i - 1].second * Base[1] + s[i - 1]) % MOD[1];
}
int siz = min(len, n);
while (sub(len - siz + 1, len) != tmpPre[siz]) siz--;
string nxt = "";
forn(i, siz, n-1) nxt += s[i];
add(nxt);
}
int n;
string s;
inline void solve()
{
basePow[0][0] = basePow[0][1] = 1;
forn(i, 1, N-1)forn(j, 0, 1) basePow[i][j] = basePow[i - 1][j] * Base[j] % MOD[j];
cin >> n;
while (n--) cin >> s, check(s);
cout << ans;
}
signed int main()
{
// MyFile
Spider
//------------------------------------------------------
// clock_t start = clock();
int test = 1;
// read(test);
// cin >> test;
forn(i, 1, test) solve();
// while (cin >> n, n)solve();
// while (cin >> test)solve();
// clock_t end = clock();
// cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
}
\[时间复杂度为:\ O(\sum_{i=1}^{n} len(s_i))
\]