HiHocoder1419 : 后缀数组四·重复旋律4&[SPOJ]REPEATS：Repeats

时间：2022-05-06 06:28:34

题面

Hihocoder

Vjudge

Sol

题目的提示说的也非常好

我对求\(LCP(P - L + len \% l, P + len \% L)\)做补充

\(len=LCP(P, P + L)\)

为什么只要求\(LCP(P - L + len \% l, P + len \% L)\)呢？

考虑在\(P - L + len \% l\)右边到\(P\)之间，它不比这里的重复次数大

考虑在\(P - L + len \% l\)左边到\(P-1\)之间，一样的它也是不增的

# include <bits/stdc++.h>

# define IL inline

# define RG register

# define Fill(a, b) memset(a, b, sizeof(a))

using namespace std;

typedef long long ll;

const int _(100010);

int n, a[_], sa[_], rk[_], y[_], height[_], t[_], vis[_];

int st[20][_], lg[_];

char s[_];

IL bool Cmp(RG int i, RG int j, RG int k){  return y[i] == y[j] && y[i + k] == y[j + k];  }

IL void Sort(){

	RG int m = 30;

	for(RG int i = 1; i <= n; ++i) ++t[rk[i] = a[i]];

	for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1];

	for(RG int i = n; i; --i) sa[t[rk[i]]--] = i;

	for(RG int k = 1; k <= n; k <<= 1){

		RG int l = 0;

		for(RG int i = n - k + 1; i <= n; ++i) y[++l] = i;

		for(RG int i = 1; i <= n; ++i) if(sa[i] > k) y[++l] = sa[i] - k;

		for(RG int i = 0; i <= m; ++i) t[i] = 0;

		for(RG int i = 1; i <= n; ++i) ++t[rk[y[i]]];

		for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1];

		for(RG int i = n; i; --i) sa[t[rk[y[i]]]--] = y[i];

		swap(rk, y); rk[sa[1]] = l = 1;

		for(RG int i = 2; i <= n; ++i) rk[sa[i]] = Cmp(sa[i - 1], sa[i], k) ? l : ++l;

		if(l >= n) break;

		m = l;

	}

	for(RG int i = 1, j = 0; i <= n; ++i){

		j = max(0, j - 1);

		while(a[j + i] == a[sa[rk[i] - 1] + j]) ++j;

		height[rk[i]] = j;

	}

}

IL int LCP(RG int xx, RG int yy){

	xx = rk[xx]; yy = rk[yy];

	if(xx > yy) swap(xx, yy);

	++xx;

	RG int l = lg[yy - xx + 1];

	return min(st[l][xx], st[l][yy - (1 << l) + 1]);

}

IL int Calc(){

	RG int ans = 0;

	for(RG int l = 1; l <= n; ++l)

		for(RG int i = 1; i + l <= n; i += l){

			RG int len = LCP(i, i + l);

			ans = max(ans, len / l + 1);

			if(i >= l - len % l) ans = max(ans, LCP(i - l + len % l, i + len % l) / l + 1);

		}

	return ans;

}

int main(RG int argc, RG char* argv[]){

	scanf(" %s", s + 1);

	n = strlen(s + 1);

	for(RG int i = 1; i <= n; ++i) a[i] = s[i] - 'a' + 1;

	for(RG int i = 2; i <= n; ++i) lg[i] = lg[i >> 1] + 1;

	Sort();

	for(RG int i = 1; i <= n; ++i) st[0][i] = height[i];

	for(RG int i = 1; i <= lg[n]; ++i)

		for(RG int j = 1; j + (1 << i) - 1 <= n; ++j)

			st[i][j] = min(st[i - 1][j], st[i - 1][j + (1 << (i - 1))]);

	printf("%d\n", Calc());

    return 0;

}

相关文章