【POJ 3693】Maximum repetition substring 重复次数最多的连续重复子串

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 100003;

int t1[N], t2[N], c[N];
void st(int *x, int *y, int *sa, int n, int m) {
	int i;
	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];
}
void mkhz(int *a, int *sa, int n, int m) {
	int *x = t1, *y = t2, *t, i, j, p;
	for(i = 0; i < n; ++i) x[i] = a[i], y[i] = i;
	st(x, y, sa, n, m);
	for(p = 1, j = 1; p < n; j <<= 1, m = p) {
		for(p = 0, i = n - j; i < n; ++i) y[p++] = i;
		for(i = 0; i < n; ++i) if (sa[i] >= j) y[p++] = sa[i] - j;
		st(x, y, sa, n, m);
		for(t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; ++i)
			x[sa[i]] = y[sa[i]] == y[sa[i - 1]] && y[sa[i] + j] == y[sa[i - 1] + j] ? p - 1 : p++;
	}
}
void mkh(int *r, int *sa, int *rank, int *h, int n) {
	int i, j, k = 0;
	for(i = 1; i <= n; ++i) rank[sa[i]] = i;
	for(i = 1; i <= n; h[rank[i++]] = k)
		for(k ? --k : 0, j = sa[rank[i] - 1]; r[i + k] == r[j + k]; ++k);
}

char s[N];
int a[N], rank[N], sa[N], h[N], n, pro = 0, f[N][30], ans, cnt, aa[N];
void mkst() {
	for(int i = 1; i <= n; ++i)
		f[i][0] = h[i];
	int k = floor(log((double)n) / log(2.0));
	for(int j = 1; j <= k; ++j)
		for(int i = 1; i <= n; ++i) {
			if (i + (1 << j) - 1 > n) break;
			f[i][j] = min(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);
		}
}
int Q(int l, int r) {
	l = rank[l]; r = rank[r];
	if (l > r) swap(l, r); ++l;
	int k = floor(log((double)(r - l + 1)) / log(2.0));
	return min(f[l][k], f[r - (1 << k) + 1][k]);
}
int main() {
	while (scanf("%s", s + 1), s[1] != ‘#‘) {
		printf("Case %d: ", ++pro);
		n = strlen(s + 1);
		for(int i = 1; i <= n; ++i) a[i] = s[i];
		mkhz(a, sa, n + 1, 130);
		mkh(a, sa, rank, h, n);
		mkst();
		ans = 0, cnt = 0;
		for(int l = 1; l < n; ++l)
			for(int i = 1; i <= n - l; i += l) {
				int k = Q(i, i + l);
				int now = k / l + 1;
				int to = i - (l - k % l);
				if (to > 0 && k % l)
					if (Q(to, to + l) >= now) ++now;
				if (now > ans) {
					aa[cnt = 1] = l;
					ans = now;
				} else if (now == ans) {
					aa[++cnt] = l;
				}
			}
		int to = 0, len = 0;
		for(int i = 1; i <= n; ++i)
			for(int j = 1; j <= cnt; ++j) {
				int k = Q(sa[i], sa[i] + aa[j]);
				if (k >= (ans - 1) * aa[j]) {
					to = sa[i];
					len = ans * aa[j];
					i = n;
					break;
				}
			}
		for(int i = to; i < to + len; ++i)
			putchar(s[i]);
		puts("");
	}
	return 0;
}