「POJ 3693」Maximum repetition substring

求一个串的最大重复次数的重复子串，要求输出字典序最小。

后缀数组，可以看罗穗骞的论文《后缀数组——处理字符串的有力工具》。

#include <cstdio>#include <cstring>#define Max(_A, _B) (_A > _B ? _A : _B)#define Min(_A, _B) (_A < _B ? _A : _B)#define R registerchar s[100010], t[100010];namespace Steaunk{    struct Suffix_Array    {        int sum[100010], SA[100010], x[100010], y[100010], H[100010], m, n, Log[100010], rmq[100010][18], rank[100010][18], P[100010][18];        bool cmp(R int i, R int j, R int k){ return x[i] == x[j] && x[i + k] == x[j + k]; }        void Swap(R int &A, R int &B){ R int t = A; A = B; B = t; }        int Query(R int l, R int r)        {            l = x[l], r = x[r];            if(l > r) Swap(l, r);            l++;            R int w = Log[r - l + 1];            return Min(rmq[l][w], rmq[r - (1 << w) + 1][w]);        }        int Least(R int l, R int r)        {            R int w = Log[r - l + 1];               return Min(rank[l][w], rank[r - (1 << w) + 1][w]);        }        int Who(R int l, R int r)        {            R int w = Log[r - l + 1];               R int t = Min(rank[l][w], rank[r - (1 << w) + 1][w]);            return t == rank[l][w] ? P[l][w] : P[r - (1 << w) + 1][w];        }        void main(char *S)        {            memset(SA, 0, sizeof(SA));            memset(x, 0, sizeof(x));            memset(sum, 0, sizeof(sum));            m = 128;            n = strlen(S + 1);            for(R int i = 1; i <= n; i++) sum[x[i] = S[i]]++;            for(R int i = 1; i <= m; i++) sum[i] += sum[i - 1];            for(R int i = n; i; i--) SA[sum[x[i]]--] = i;            for(R int i = 1; i < n; i <<= 1)            {                R int pos = 0;                for(R int j = n - i + 1; j <= n; j++) y[++pos] = j;                for(R int j = 1; j <= n; j++) if(SA[j] > i) y[++pos] = SA[j] - i;                for(R int j = 0; j <= m; j++) sum[j] = 0;                for(R int j = 1; j <= n; j++) sum[x[j]]++;                for(R int j = 1; j <= m; j++) sum[j] += sum[j - 1];                for(R int j = n; j; j--) SA[sum[x[y[j]]]--] = y[j];                y[SA[1]] = 1;                for(R int j = 2; j <= n; j++) y[SA[j]] = y[SA[j - 1]] + !cmp(SA[j], SA[j - 1], i);                for(R int j = 1, t = x[j]; j <= n; x[j] = y[j], y[j++] = t);                m = x[SA[n]];                if(m == n) break;            }            memset(H, 0, sizeof(H));            for(R int i = 1; i <= n; i++)            {                H[x[i]] = Max(H[x[i - 1]] - 1, 0);                while(S[i + H[x[i]]] == S[SA[x[i] - 1] + H[x[i]]]) H[x[i]]++;            }            for(R int i = 2; i <= n; i++) Log[i] = Log[i >> 1] + 1;            memset(rmq, 0, sizeof(rmq));            memset(rank, 0, sizeof(rank));            for(R int i = 1; i <= n; i++) rmq[i][0] = H[i];            for(R int j = 1; j < 18; j++)                for(R int i = 1; i <= n; i++)                    rmq[i][j] = Min(rmq[i][j - 1], rmq[i + (1 << (j - 1))][j - 1]);        }        void Extra()        {            for(R int i = 1; i <= n; i++) rank[i][0] = x[i], P[i][0] = i;            for(R int j = 1; j < 18; j++)                for(R int i = 1; i <= n; i++)                    rank[i][j] = Min(rank[i][j - 1], rank[i + (1 << (j - 1))][j - 1]),                    P[i][j] = (rank[i][j] == rank[i][j - 1] ?                                P[i][j - 1] : P[i + (1 << (j - 1))][j - 1]);        }    } a, b;     int Ans, L, u;    void main()    {        R int n = strlen(s + 1); L = 1, Ans = 1, u = 1;         for(R int i = 1; i <= n; i++) if(s[i] < s[L]) L = i;        a.main(s), a.Extra();        for(R int i = 1; i <= n; i++) t[i] = s[n - i + 1]; t[n] = '\0';        b.main(t);        for(R int i = 1; i <= n; i++)               for(R int j = i; j <= n; j += i)            {                if(s[j] != s[j - i]) continue;                R int t1 = a.Query(j + 1 - i, j + 1), t2 = b.Query(n - j + 2, n - j + 2 + i);                R int K = (t1 + t2 + 1) / i + 1, v = t1 + t2 + 1 + i - K * i;                if(Ans < K) Ans = K, L = a.Who(j - t2 - i, j - t2 - i + v), u = i;                else if(Ans == K && a.x[L] > a.Least(j - t2 - i, j - t2 - i + v))                     L = a.Who(j - t2 - i, j - t2 - i + v), u = i;            }        for(R int i = 1; i <= Ans; i++)            for(R int j = 1; j <= u; j++)                printf("%c", s[L++]);        puts("");    }}int main(){    R int cnt = 0;    while(scanf("%s", s + 1), s[1] != '#')     {        printf("Case %d: ", ++cnt);        Steaunk::main();    }    return 0;}