【洛谷3546_BZOJ2803】[POI2012]PRE-Prefixuffix（String Hash）

Problem:

洛谷3546

Analysis：

I gave up and saw other's solution when I had nearly thought of the method ... What a pity

Let's define a border of string \(s\) as a prefix \(p\) of \(s\) that \(p\) is also a suffix of \(s\). What the problem asked us to look for is a number \(L\), that the prefix of length \(L\) can be divided into two string \(s1\) and \(s2\) , and the suffix of length \(L\) can be divided into two string \(s2\) and \(s1\), so that this pair of prefix and suffix is cyclically equivalent. Obviously, \(s1\) is a border of string \(s\).

(not the end...

Code:

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cctype>
using namespace std;

namespace zyt
{
    template<typename T>
    inline bool read(T &x)
    {
        char c;
        bool f = false;
        x = 0;
        do
            c = getchar();
        while (c != EOF && c != '-' && !isdigit(c));
        if (c == EOF)
            return false;
        if (c == '-')
            f = true, c = getchar();
        do
            x = x * 10 + c - '0', c = getchar();
        while (isdigit(c));
        if (f)
            x = -x;
        return true;
    }
    inline bool read(char *const s)
    {
        return ~scanf("%s", s);
    }
    template<typename T>
    inline void write(T x)
    {
        static char buf[20];
        char *pos = buf;
        if (x < 0)
            putchar('-'), x = -x;
        do
            *pos++ = x % 10 + '0';
        while (x /= 10);
        while (pos > buf)
            putchar(*--pos);
    }
    const int N = 1e6 + 10;
    int f[N], n;
    // f[i] is the maximum length of the border of substr[i, n - i - 1]
    char str[N];
    namespace Hash
    {
        typedef long long ll;
        typedef pair<int, int> pii;
        typedef pii hash_t;
        hash_t h[N], pow[N];
        const hash_t seed = hash_t(61, 67), p = hash_t(1e9 + 7, 1e9 + 9);
        hash_t operator + (const hash_t &a, const hash_t &b)
        {
            return hash_t((a.first + b.first) % p.first, (a.second + b.second) % p.second);
        }
        hash_t operator - (const hash_t &a, const hash_t &b)
        {
            return hash_t((a.first - b.first + p.first) % p.first, 
                (a.second - b.second + p.second) % p.second);
        }
        hash_t operator * (const hash_t &a, const hash_t &b)
        {
            return hash_t(int((ll)a.first * b.first % p.first), 
                int((ll)a.second * b.second % p.second));
        }
        void init()
        {
            pow[0] = make_pair(1, 1);
            for (int i = 1; i < N; i++)
                pow[i] = pow[i - 1] * seed;
        }
        inline int ctoi(const char c)
        {
            return c - 'a';
        }
        void get(const char *const s)
        {
            h[0] = make_pair(ctoi(s[0]), ctoi(s[0]));
            for (int i = 1; i < n; i++)
                h[i] = h[i - 1] * seed + make_pair(ctoi(s[i]), ctoi(s[i]));
        }
        hash_t extract(const int l, const int r)
        {
            return l ? (h[r] - h[l - 1] * pow[r - l + 1]) : h[r];
        }
    }
    using namespace Hash;
    void mk_f()
    {
        f[n >> 1] = 0;
        for (int i = (n >> 1) - 1; i >= 0; i--)
        {
            f[i] = min(f[i + 1] + 2, (n >> 1) - i);
            while (f[i] && extract(i, i + f[i] - 1) != extract(n - i - f[i], n - i - 1))
                --f[i];
        }
    }
    int work()
    {
        read(n), read(str);
        init();
        get(str);
        mk_f();
        int ans = 0;
        for (int i = 1; i <= (n >> 1); i++)
            if (extract(0, i - 1) == extract(n - i, n - 1))
                ans = max(ans, i + f[i]);
        write(ans);
        return 0;
    }
}
int main()
{
    return zyt::work();
}