后缀数组应用——回文子串

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最长回文子串 (ural1297)

给出一个字符串 S，求出它的最长回文子串是什么。

当然，可以用 manacher 做，但是这篇要说的的是用后缀数组的做法：枚举每一位，求以这一位为中心的最长回文子串是什么。我们把原字符串反转并把它接在原字符串的后面，中间加入一个分隔符，这样问题就变成了求这个新的字符串的某两个后缀的最长公共前缀（如图）。注意回文子串为偶数和为奇数是两种情况，要区分一下。

总的时间复杂度是 O( nlogn )，如果RMQ 用 O( n ) 的方法预处理，就能做到 O( n )。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAX_N = 1005;int a[MAX_N << 1], r[MAX_N << 1], h[MAX_N << 1], n, l, sa[MAX_N << 1];int ws[MAX_N << 1], wv[MAX_N << 1], wa[MAX_N << 1], wb[MAX_N << 1];char s[MAX_N];int rmp[33][MAX_N << 1], lg[MAX_N << 1];void da(int *a, int *sa, int n, int m){int *x = wa, *y = wb;for (int i = 0; i < m; i ++) ws[i] = 0;for (int i = 0; i < n; i ++) ws[x[i] = a[i]] ++;for (int i = 1; i < m; i ++) ws[i] += ws[i - 1];for (int i = n - 1; i >= 0; i --) sa[-- ws[x[i]]] = i;for (int k = 1; k <= n; k <<= 1){int p = 0;for (int i = n - k; i < n; i ++)  y[p ++] = i;for (int i = 0; i < n; i ++) if (sa[i] >= k) y[p ++] = sa[i] - k;for (int i = 0; i < n; i ++) wv[i] = x[y[i]];for (int i = 0; i < m; i ++) ws[i] = 0;for (int i = 0; i < n; i ++) ws[wv[i]] ++;for (int i = 1; i < m; i ++) ws[i] += ws[i - 1];for (int i = n - 1; i >= 0; i --) sa[-- ws[wv[i]]] = y[i];swap(x, y); p = 1; x[sa[0]] = 0;for (int i = 1; i < n; i ++) x[sa[i]] = (y[sa[i - 1]] == y[sa[i]]) && (y[sa[i - 1] + k] == y[sa[i] + k]) ? p - 1 : p ++;if (p >= n) break; m = p;}}void calc(){int k = 0, j;for (int i = 1; i <= n; i ++) r[sa[i]] = i;for (int i = 0; i < n; h[r[i ++]] = k)for (k ? k -- : 0, j = sa[r[i] - 1]; a[i + k] == a[j + k]; k ++);}void RMQ(){lg[0] = -1;for (int i = 1; i <= n; i ++) lg[i] = (i & (i - 1)) == 0 ? lg[i - 1] + 1 : lg[i - 1];for (int i = 1; i <= n; i ++) rmp[0][i] = i;for (int i = 1; i <= lg[n]; i ++)for (int j = 1; j + ((1 << i) - 1) <= n; j ++){int a = rmp[i - 1][j], b = rmp[i - 1][j + (1 << (i - 1))];if (h[a] < h[b]) rmp[i][j] = a;else rmp[i][j] = b;}}int ask(int a, int b){int k = lg[b - a + 1]; b -= (1 << k) - 1;a = rmp[k][a], b = rmp[k][b];return h[a] < h[b] ? a : b;}int lcp(int a, int b){a = r[a]; b = r[b];if (a > b) swap(a, b);return h[ask(a + 1, b)];}void init(){scanf("%s", s); l = strlen(s);for (int i = 0; i < l; i ++) a[i] = (int)s[i];a[l] = 1;for (int i = 0; i < l; i ++) a[i + l + 1] = a[l - i - 1];n = (l << 1) + 1; a[n] = 0; da(a, sa, n + 1, 128); calc(); RMQ();}void doit(){int ans = 0, st;for (int i = 0; i < l; i ++){int k = lcp(i, n - i);if (2 * k > ans) ans = 2 * k, st = i - k;k = lcp(i, n - i - 1);if (2 * k - 1 > ans) ans = 2 * k - 1, st = i - k + 1;}for (int i = st; i <= st + ans - 1; i ++) printf("%c", s[i]);}int main(){init();doit();return 0;}

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