HUST 1352 Repetitions of Substrings

Description

The “repetitions” of a string S(whose length is n) is a maximum number “k” such that: 1) k is a factor of n 2) S[0..n/k-1] = S[p*(n/k)..(p+1)*(n/k)-1] for all that (1 <= p < n/k) for example: the repetitions of “aaaaaa”is 6. the repetitions of “abababab”is 4. the repetitions of “abcdef”is 1. Now, given a string S and a number K, please tell me how many substrings of S have repetitions NOT less than K.

Input

The input consists of several instances, each one for a single line. S K S is a string, K is a number. Check the Description for their meanings. S contains lowercase letters(ie 'a'..'z') only. 1 <= length of S <= 100000. 1 <= K <= length of S.

Output

For each instance, output the number of substring whose repetitions is NOT less than K.

Sample Input

abcabc 2
acmac 3
      

Sample Output

1

0

判断有多少子串满足循环至少k次，用后缀数组，计算出height

然后循环对于每个位置判断可能的最大长度,除k，得到循环节的可能长度，分别枚举每个循环节判断。

这题数据很水，让很多有问题的程序跑过了，错误点见注释。


       
#include<set>
#include<map>
#include<ctime>
#include<cmath>
#include<stack>
#include<queue>
#include<bitset>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define per(i,j,k) for (int i = j; i >= k; i--)
#define lson x << 1, l, mid
#define rson x << 1 | 1, mid + 1, r
#define fi first
#define se second
#define mp(i,j) make_pair(i,j)
#define pii pair<int,int>
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const double eps = 1e-8;
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int N = 1e5 + 10;
const int read()
{
	char ch = getchar();
	while (ch<'0' || ch>'9') ch = getchar();
	int x = ch - '0';
	while ((ch = getchar()) >= '0'&&ch <= '9') x = x * 10 + ch - '0';
	return x;
}
int m;

struct Sa
{
	char s[N];
	int rk[2][N], sa[N], h[N], w[N], now, n;
	int rmq[N][20], lg[N], f[N];

	bool GetS()
	{
		return scanf("%s%d", s + 1, &m) != EOF;
	}

	void getsa(int z, int &m)
	{
		int x = now, y = now ^= 1;
		rep(i, 1, z) rk[y][i] = n - i + 1;
		for (int i = 1, j = z; i <= n; i++)
			if (sa[i] > z) rk[y][++j] = sa[i] - z;

		rep(i, 1, m) w[i] = 0;
		rep(i, 1, n) w[rk[x][rk[y][i]]]++;
		rep(i, 1, m) w[i] += w[i - 1];
		per(i, n, 1) sa[w[rk[x][rk[y][i]]]--] = rk[y][i];
		for (int i = m = 1; i <= n; i++)
		{
			int *a = rk[x] + sa[i], *b = rk[x] + sa[i - 1];
			rk[y][sa[i]] = *a == *b&&*(a + z) == *(b + z) ? m - 1 : m++;
		}
	}

	void getsa(int m)
	{
		rk[1][0] = now = sa[0] = s[0] = 0;
		n = strlen(s + 1);
		rep(i, 1, m) w[i] = 0;
		rep(i, 1, n) w[s[i]]++;
		rep(i, 1, m) rk[1][i] = rk[1][i - 1] + (bool)w[i];
		rep(i, 1, m) w[i] += w[i - 1];
		rep(i, 1, n) rk[0][i] = rk[1][s[i]];
		rep(i, 1, n) sa[w[s[i]]--] = i;

		rk[1][n + 1] = rk[0][n + 1] = 0;	//多组的时候容易出bug
		for (int x = 1, y = rk[1][m]; x <= n && y <= n; x <<= 1) getsa(x, y);
		for (int i = 1, j = 0; i <= n; h[rk[now][i++]] = j ? j-- : 0)
		while (rk[now][i] - 1 && s[sa[rk[now][i] - 1] + j] == s[i + j]) ++j;
	}

	void getrmq()
	{
		h[n + 1] = h[1] = lg[1] = 0;
		rep(i, 2, n) rmq[i][0] = h[i], lg[i] = lg[i >> 1] + 1;
		for (int i = 1; (1 << i) <= n; i++)
		{
			rep(j, 2, n)
			{
				if (j + (1 << i) > n + 1) break;
				rmq[j][i] = min(rmq[j][i - 1], rmq[j + (1 << i - 1)][i - 1]);
			}
		}
	}

	int lcp(int x, int y)
	{
		int l = min(rk[now][x], rk[now][y]) + 1, r = max(rk[now][x], rk[now][y]);
		return min(rmq[l][lg[r - l + 1]], rmq[r - (1 << lg[r - l + 1]) + 1][lg[r - l + 1]]);
	}

	void work()
	{
		getsa(255);
		if (m == 1)
		{
			printf("%lld\n", 1LL * n * (n + 1) / 2);
			return;
		}
		getrmq();
		
		LL ans = 0;
		rep(i, 1, n) f[i] = 0;
		rep(i, 1, n)
		{
			int t = max(h[i], h[i + 1]) / (m - 1);
			rep(L, 1, t)
			{
				if (sa[i] + L > n) break;
				if (f[L] == i) continue;

				int R = lcp(sa[i], sa[i] + L);
				if (R / L  + 2 >  m) ans += R / L + 2 - m;
				rep(k, 1, R / L) f[k*L] = i;
				/*
					if (R >= L) break;
					这里有毒,加上这句话可以快很多的AC，然而这是错的，看了很多人的程序，都是这么写的。
					明显的样例是 aabaab 2    
					正确答案是 3 而加上这句话会输出 2 明显是不对的，只能说数据很水。
				*/
			}
		}
		printf("%lld\n", ans);
	}
}sa;

int main()
{
	while (sa.GetS()) sa.work();
	return 0;
}