POJ-3693/HDU-2459 Maximum repetition substring 最多重复次数的子串（需要输出具体子串，按字典序）

#include <cstdio>
#include <string>
#include <iostream>
#include <set>
#include <algorithm>
#include <vector>
#include <map>
#include <cstring>
#include <queue>
#include <cmath>
#define LL int
using namespace std;
const LL N = 100055;
string s;
int lx[N];
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;//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)
    {
        m=1050;
        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();
    }
    void solve()//cascnacnacdsdaj
    {
        int ans = 2,al,st=-1;
        int cnt=0;
        set<char>fuck;
        for(int i=0; i<slen; i++)
            fuck.insert(s[i]);
        for (int len = 1; len <= slen; len++)
        {
            for (int i = 0; i+len< slen; i+=len)
            {
                int r1 = Rank[i], r2 = Rank[i+ len];
                int lcp = query(r1, r2);
                int sta=i;
                if(ans<=lcp /  len + 1)
                {
                    if(ans<lcp /  len + 1)
                    {
                        cnt=0;
                        lx[cnt++]=len;
                    }
                    else lx[cnt++]=len;
                    ans=lcp /  len + 1;
                }

                if (i - len + lcp%len >= 0)
                {
                    sta=i - len + lcp%len;
                    if(sta==i-len)sta++;
                    lcp = query(Rank[sta], Rank[sta+len]);
                    if(ans<=lcp /  len + 1)
                    {
                        if(ans<lcp /  len + 1)
                        {
                            cnt=0;
                            lx[cnt++]=len;
                        }
                        else lx[cnt++]=len;
                        ans=lcp /  len + 1;
                    }
                }
            }
        }
        int pos=-1,lxl;
        for(int i=1; i<=slen; i++)
        {
            for(int j=0;j<cnt;j++)
            {
                int sta=SA[i];
                if(sta+lx[j]>=slen)break;
                int rk=Rank[sta+lx[j]];
                int lcp = query(i, rk);
                if(ans==lcp / lx[j] + 1)
                {
                    for(int i=0; i<ans*lx[j]; i++)
                        cout<<s[i+sta];
                    cout<<endl;
                    goto li;
                }
            }
        }
        cout<<*fuck.begin()<<endl;
        return;
li:
        ;

    }
} sf;
int main()
{
    cin.sync_with_stdio(false);
    //freopen("in.txt","r",stdin);
    //freopen("o1.txt","w",stdout);
    int cas=1;
    while (cin >> s)
    {
        if(s=="#")break;
        cout<<"Case "<<cas++<<": ";
        sf.init(s);
        sf.solve();
    }
    return 0;
}