「POJ 3693」Maximum repetition substring

后缀数组详解

最新推荐文章于 2021-04-14 21:15:43 发布

原创最新推荐文章于 2021-04-14 21:15:43 发布 · 381 阅读

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本文介绍了一种处理字符串问题的有效工具——后缀数组，并通过解决一个具体问题来展示其使用方法。该问题要求找出一个字符串中最大重复次数的重复子串，并要求输出字典序最小的子串。文章提供了一个详细的后缀数组实现示例，包括如何构建后缀数组、如何利用后缀数组进行查询等。

题意：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》。

来自《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 register
char 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;
}