NOI2016优秀的拆分

• 一种想法是枚举分割位置， 然后考虑前面部分有多少种可行的AA拆分方式， 后面部分有多少种可行的BB拆分方式， 然后乘法原理即可
• 那么问题是如何快速求出合法方案
• 解法是首先枚举长度len， 然后将序列分成$\frac{n}{len}$段， 然后我们对于每个连续的三个段i，j, k， 求一下i, j的最长公共后缀a， j，k的最长公共前缀b,
• 如果a + b < len， 显然无法构造
• 如果a + b >= len， 那么整整一个区间都可以构造出来， 差分统计一下答案
• 因为某些地方没清空以及某些地方特判写狗了， WA成狗

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#define ll long long
#define M 60010
using namespace std;
int read() {
	int nm = 0, f = 1;
	char c = getchar();
	for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
	return nm * f;
}
int log[M];
struct SA {
	char s[M];
	int sa[M], tmp[M], tex[M], rnk[M], f[M][17], n, m, q;

	bool check(int i, int j, int w) {
		return tmp[i] == tmp[j] && tmp[i + w] == tmp[j + w];
	}

	void qsort() {
		for(int i = 0; i <= m; i++) tex[i] = 0;
		for(int i = 1; i <= n; i++) tex[rnk[tmp[i]]]++;
		for(int i = 1; i <= m; i++) tex[i] += tex[i - 1];
		for(int i = n; i >= 1; i--) sa[tex[rnk[tmp[i]]]--] = tmp[i];
	}

	void suffix() {
		memset(tmp, 0, sizeof(tmp));
		memset(tex, 0, sizeof(tex));
		memset(sa, 0, sizeof(sa));
		memset(rnk, 0, sizeof(rnk));
		memset(f, 0, sizeof(f));
		m = 127, q = 0;
		for(int i = 1; i <= n; i++) rnk[i] = s[i], tmp[i] = i;
		qsort();
		for(int w = 1; q < n; w <<= 1, m = q) {
			q = 0;
			for(int i = n - w + 1; i <= n; i++) tmp[++q] = i;
			for(int i = 1; i <= n; i++) if(sa[i] > w) tmp[++q] = sa[i] - w;
			qsort();
			swap(rnk, tmp);
			rnk[sa[1]] = q = 1;
			for(int i = 2; i <= n; i++) if(check(sa[i - 1], sa[i], w)) rnk[sa[i]] = q;
				else rnk[sa[i]] = ++q;
		}
		q = 0;
	//	cerr << "\n";
		for(int i = 1; i <= n; i++) {
			if(q) q--;
			int j = sa[rnk[i] + 1];
			if(rnk[i] == n) continue;
			while(s[i + q] == s[j + q] && i + q <= n && j + q <= n) q++;
			f[rnk[i]][0] = q;
		//	cerr << q << " ";
			//if(q >n) cerr<<"!";
		}
	//	cerr << "\n";
		//for(int i = 1; i <= n; i++) cerr << rnk[i] << " ";
	//	cout << "\n";
		for(int j = 1; j <= 15; 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 lcp(int a, int b) {
		a = rnk[a], b = rnk[b];
		if(a > b) swap(a, b);
		int k = log[b - a];
		return min(f[a][k], f[b - (1 << k)][k]);
	}
} a, b;
int be[M], ed[M];
char s[M];

void work() {
	memset(s, 0, sizeof(s));
	scanf("%s", s + 1);
	int n = strlen(s + 1);
	for(int i = 1; i <= n; i++) {
		a.s[i] = s[i], b.s[i] = s[n - i + 1];
	}
	memset(be, 0, sizeof(be));
	memset(ed, 0, sizeof(ed));
	a.n = b.n = n;
	a.suffix();
	b.suffix();
	for(int len = 1; len <= n / 2; len++) {
		for(int i = len, j = i + len; j <= n; i += len, j += len) {
			int x = min(a.lcp(i, j), len), y = min(b.lcp(n - (i - 1) + 1, n - (j - 1) + 1), len - 1);
			int t = x + y - len + 1;
			if(x + y >= len) {
				be[i - y]++;
				be[i - y + t]--;
				ed[j + x - t]++;
				ed[j + x]--;
			}
		}
	}
	ll ans = 0;
	for(int i = 1; i <= n; i++) be[i] += be[i - 1], ed[i] += ed[i - 1], ans += 1ll * ed[i - 1] * be[i];
	cout << ans << "\n";
}

int main() {
	for(int i = 2; i < M; i++) log[i] = log[i >> 1] + 1;
	int T = read();
	while(T--) work();
	return 0;
}

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*/