后缀数组

DA算法

/* * suffix array * 倍增算法 O(n*logn) * 待排序数组长度为n,放在0~n-1中,在最后面补一个0 * da(str, sa, rank, height, n, m); * 例如: * n = 8; * num[] = { 1, 1, 2, 1, 1, 1, 1, 2, $ }; 注意num最后一位为0,其他大于0 * rank[] = { 4, 6, 8, 1, 2, 3, 5, 7, 0 }; rank[0~n-1]为有效值,rank[n]必定为0无效值 * sa[] = { 8, 3, 4, 5, 0, 6, 1, 7, 2 }; sa[1~n]为有效值,sa[0]必定为n是无效值 * height[]= { 0, 0, 3, 2, 3, 1, 2, 0, 1 }; height[2~n]为有效值 * 稍微改动可以求最长公共前缀，需要注意两串起始位置相同的情况 * 另外需要注意的是部分数组需要开两倍空间大小 */
const int MAXN = 20010;

int t1[MAXN];
int t2[MAXN];
int c[MAXN];    // 求SA数组需要的中间变量,不需要赋值

// 待排序的字符串放在s数组中,从s[0]到s[n-1],长度为n,且最大值小于m,
// 除s[n-1]外的所有s[i]都大于0,r[n-1]=0
// 函数结束以后结果放在sa数组中
bool cmp(int *r, int a, int b, int l)
{
    return r[a] == r[b] && r[a + l] == r[b + l];
}

void da(int str[], int sa[], int rank[], int height[], int n, int m)
{
    n++;
    int i, j, p, *x = t1, *y = t2;  // 第一轮基数排序,如果s的最大值很大,可改为快速排序
    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;
        // 直接利用sa数组排序第二关键字
        for (i = n - j; i < n; i++)
        {
            y[p++] = i;                 // 后面的j个数第二关键字为空的最小
        }
        for (i = 0; i < n; i++)
        {
            if (sa[i] >= j)
            {
                y[p++] = sa[i] - j;     // 这样数组y保存的就是按照第二关键字排序的结果
            }
        }
        // 基数排序第一关键字
        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];    // 根据sa和x数组计算新的x数组
        }
        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;                          // 下次基数排序的最大值
    }
    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;
    }
}

int _rank[MAXN], height[MAXN];
int RMQ[MAXN];
int mm[MAXN];

int best[20][MAXN];

void initRMQ(int n)
{
    mm[0] = -1;
    for (int i = 1; i <= n; i++)
    {
        mm[i] = ((i & (i - 1)) == 0) ? mm[i - 1] + 1 : mm[i - 1];
    }
    for (int i = 1; i <= n; i++)
    {
        best[0][i] = i;
    }
    for (int i = 1; i <= mm[n]; i++)
    {
        for (int j = 1; j + (1 << i) - 1 <= n; j++)
        {
            int a = best[i - 1][j];
            int b = best[i - 1][j + (1 << (i - 1))];
            if (RMQ[a] < RMQ[b])
            {
                best[i][j] = a;
            }
            else
            {
                best[i][j] = b;
            }
        }
    }
}

int askRMQ(int a, int b)
{
    int t;
    t = mm[b - a + 1];
    b -= (1 << t) - 1;
    a = best[t][a];
    b = best[t][b];
    return RMQ[a] < RMQ[b] ? a : b;
}

int lcp(int a, int b)
{
    a = _rank[a];
    b = _rank[b];
    if (a > b)
    {
        swap(a,b);
    }
    return height[askRMQ(a + 1, b)];
}

char str[MAXN];
int r[MAXN];
int sa[MAXN];

int main()
{
    while (scanf("%s", str) == 1)
    {
        int len = (int)strlen(str);
        int n = 2 * len + 1;
        for (int i = 0; i < len; i++)
        {
            r[i] = str[i];
        }
        for (int i = 0; i < len; i++)
        {
            r[len + 1 + i] = str[len - 1 - i];
        }
        r[len] = 1;
        r[n] = 0;
        da(r, sa, _rank, height, n, 128);
        for (int i = 1; i <= n; i++)
        {
            RMQ[i]=height[i];
        }
        initRMQ(n);
        int ans = 0, st = 0;
        int tmp;
        for (int i = 0; i < len; i++)
        {
            tmp = lcp(i, n - i);    // 偶对称
            if (2 * tmp > ans)
            {
                ans = 2 * tmp;
                st = i - tmp;
            }
            tmp=lcp(i, n - i - 1);  // 奇数对称
            if (2 * tmp - 1 > ans)
            {
                ans = 2 * tmp - 1;
                st = i - tmp + 1;
            }
        }
        str[st + ans] = 0;
        printf("%s\n", str + st);
    }
    return 0;
}

DC3算法

da[]和str[]数组都要开大三倍，相关数组也是三倍

/*
 * 后缀数组
 * DC3算法,复杂度O(n)
 * 所有的相关数组都要开三倍 
 */
const int MAXN = 2010;
#define F(x) ((x) / 3 + ((x) % 3 == 1 ? 0 : tb))
#define G(x) ((x) < tb ? (x) * 3 + 1 : ((x)-tb) * 3 + 2)

int wa[MAXN * 3], wb[MAXN * 3], wv[MAXN * 3], wss[MAXN * 3];

int c0(int *r, int a, int b)
{
    return r[a] == r[b] && r[a + 1] == r[b + 1] && r[a + 2] == r[b + 2];
}

int c12(int k, int *r, int a, int b)
{
    if(k == 2)
    {
        return r[a] < r[b] || (r[a] == r[b] && c12(1, r, a + 1, b + 1));
    }
    else
    {
        return r[a] < r[b] || (r[a] == r[b] && wv[a + 1] < wv[b + 1]);
    }
}

void sort(int *r, int *a, int *b, int n, int m)
{
    int i;
    for (i = 0; i < n; i++)
    {
        wv[i] = r[a[i]];
    }
    for (i = 0; i < m; i++)
    {
        wss[i] = 0;
    }
    for (i = 0; i < n; i++)
    {
        wss[wv[i]]++;
    }
    for (i = 1; i < m; i++)
    {
        wss[i] += wss[i - 1];
    }
    for (i = n - 1; i >= 0; i--)
    {
        b[--wss[wv[i]]] = a[i];
    }
}

void dc3(int *r, int *sa, int n, int m)
{
    int i, j, *rn = r + n;
    int *san = sa + n, ta = 0, tb = (n+1)/3, tbc = 0, p;
    r[n] = r[n+1] = 0;
    for (i = 0; i < n; i++)
    {
        if (i % 3 != 0)
        {
            wa[tbc++] = i;
        }
    }
    sort(r + 2, wa, wb, tbc, m);
    sort(r + 1, wb, wa, tbc, m);
    sort(r, wa, wb, tbc, m);
    for (p = 1, rn[F(wb[0])] = 0, i = 1; i < tbc; i++)
    {
        rn[F(wb[i])] = c0(r, wb[i - 1], wb[i]) ? p - 1 : p++;
    }
    if (p < tbc)
    {
        dc3(rn, san, tbc, p);
    }
    else
    {
        for (i = 0; i < tbc; i++)
        {
            san[rn[i]] = i;
        }
    }
    for (i = 0; i < tbc; i++)
    {
        if (san[i] < tb)
        {
            wb[ta++] = san[i] * 3;
        }
    }
    if (n % 3 == 1)
    {
        wb[ta++] = n - 1;
    }
    sort(r, wb, wa, ta, m);
    for (i = 0; i < tbc; i++)
    {
        wv[wb[i] = G(san[i])] = i;
    }
    for (i = 0, j = 0, p = 0; i < ta && j < tbc; p++)
    {
        sa[p] = c12(wb[j] % 3, r, wa[i], wb[j]) ? wa[i++] : wb[j++];
    }
    for (; i < ta; p++)
    {
        sa[p] = wa[i++];
    }
    for (; j < tbc; p++)
    {
        sa[p] = wb[j++];
    }
}

//  str和sa也要三倍
void da(int str[], int sa[], int rank[], int height[], int n,int m)
{
    for (int i = n; i < n * 3; i++)
    {
        str[i] = 0;
    }
    dc3(str, sa, n+1, m);
    int i, j, k = 0;
    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;
    }
}