CF1200E Compress Words 題解

Athanasy發表於2024-04-06

題目連結：CF 或者 洛谷

注意到總字串長度不超過 \(1e6\)，對於兩個串之間找前字尾匹配，只要能暴力列舉長度，\(check\ 為\ O(1)\)，那麼最後顯然線性複雜度。可以考慮 \(kmp\)，也可以考慮字串雜湊，最好上雙雜湊，然後拼串顯然是在尾部繼續新增新的字首雜湊，這個需要新增的串可以單獨由匹配長度拿出來。匹配長度顯然由最長開始列舉，直到答案串尾部的雙雜湊與待匹配串的字首雙雜湊相同為止，空串特判。

參照程式碼 
#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'
//用於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;
}

\[時間複雜度為:\ O(\sum_{i=1}^{n} len(s_i)) \]

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