hihocoder-1419 后缀数组四·重复旋律4 求连续重复次数最多的子串

Posted 2020-10-07 Luke_Ye

对于重复次数，如果确定了重复子串的长度len，那重复次数k=lcp(start,start+len)/len+1。而暴力枚举start和len的复杂度是O(n^2)，不能接受。而有一个规律，若我们只枚举len的整数倍作为起始，如果将它向前移动小于len位可以构成重复次数更长的串，那么那个位置p=start-len+lcp%len。所以每次我们计算两者并求max再与ans做max即可。

#include <cstdio>
#include <string>
#include <iostream>
#include <set>
#include <algorithm>
#include <vector>
#include <map>
#include <cstring>
#include <queue>
#define LL int
using namespace std;
const LL N = 100055;

class SF
{
    //N:数组大小
public:
    int x[N], y[N], c[N];
    int Height[N], str[N], SA[N], Rank[N];//Height数组从2开始,SA记录Rank=i的下标
    int slen;
    int m = 1050;//字符集处理大小(传入如果不是数字，需要做位移转换)
    bool cmp(int* r, int a, int b, int l) {
        return r[a] == r[b] && r[a + l] == r[b + l];
    }
    void Suffix(int n) {
        ++n;
        int i, j, p;
        for (i = 0; i < m; ++i) c[i] = 0;
        for (i = 0; i < n; ++i) c[x[i] = str[i]]++;
        for (i = 1; i < m; ++i) c[i] += c[i - 1];
        for (i = n - 1; i >= 0; --i) SA[--c[x[i]]] = i;
        for (j = 1; j <= n; j <<= 1) {
            p = 0;
            for (i = n - j; i < n; ++i) y[p++] = i;
            for (i = 0; i < n; ++i) if (SA[i] >= j) y[p++] = SA[i] - j;
            for (i = 0; i < m; ++i) c[i] = 0;
            for (i = 0; i < n; ++i) c[x[y[i]]]++;

            for (i = 1; i < m; ++i) c[i] += c[i - 1];
            for (i = n - 1; i >= 0; --i) SA[--c[x[y[i]]]] = y[i];

            swap(x, y);
            p = 1; x[SA[0]] = 0;
            for (i = 1; i < n; ++i) {
                x[SA[i]] = cmp(y, SA[i - 1], SA[i], j) ? p - 1 : p++;
            }
            if (p >= n)break;
            m = p;
        }

        int k = 0;
        n--;
        for (i = 0; i <= n; ++i) Rank[SA[i]] = i;
        for (i = 0; i < n; ++i) {
            if (k)--k;
            j = SA[Rank[i] - 1];
            while (str[i + k] == str[j + k])++k;
            Height[Rank[i]] = k;
            //cout << k << endl;
        }
    }
    static const int bitlen = 25;
    LL lg2(LL p)//计算log2(n)
    {
        return (LL)(log(p) / log(2));
    }
    LL dp[bitlen][N];
    LL bit[bitlen];
    void initRMQ()//初始化
    {
        bit[0] = 1;
        for (int i = 1; i < bitlen; i++) bit[i] = 2 * bit[i - 1];
        for (int i = 0; i <= slen; i++)
            dp[0][i] = Height[i];
        dp[0][0] = dp[0][1] = 0;
        for (LL i = 1; bit[i] < slen + 1; i++)
            for (LL j = 0; j + bit[i] <= slen + 1; j++)
                dp[i][j] = min(dp[i - 1][j], dp[i - 1][j + bit[i - 1]]);
    }
    LL query(LL l, LL r)//查询两个Rank之间的lcp
    {
        if (r == l) return slen - SA[l];
        if (l > r) swap(l, r);
        l++;
        LL mig = lg2(r - l + 1.0);
        return min(dp[mig][l], dp[mig][r - bit[mig] + 1]);
    }
    void init(string &s)
    {
        slen = s.size();
        for (int i = 0; i < slen; i++)
            str[i] = s[i] - ‘a‘ + 2;//如果是字符，映射成从1开始的序列
        str[slen] = 1;//1作为结束符，防止越界
        Suffix(slen);
        initRMQ();
    }
    int solve()
    {
        int ans = 0;
        for (int len = 1; len <= slen; len++)
        {
            for (int i = 0; i < slen; i+=len)
            {
                int r1 = Rank[i], r2 = Rank[i+ len];
                int lcp = query(r1, r2);
                ans = max(ans, lcp / len + 1);
                if (i - len + lcp%len >= 0)
                {
                    lcp = query(Rank[i - len + lcp%len], Rank[i + lcp%len]);
                    ans = max(ans, lcp /  len + 1);
                }    
            }
        }
        return ans;
    }
}sf;
int main()
{
    cin.sync_with_stdio(false);
    string s;
    while (cin >> s)
    {
        sf.init(s);
        cout << sf.solve() << endl;
    }
    return 0;
}