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;
}
}
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