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

时间：2024-01-20 18:45:33

后缀数组的论文里的例题，论文里的题解并没有看懂，，，

求一个重复次数最多的连续重复子串，又因为要找最靠前的，所以扫的时候记录最大的重复次数为$ans$，扫完后再后从头暴力扫到尾找重复次数为$ans$的第一个子串的开头，break输出就可以了

#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;

}

终于A了233

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