【BZOJ 4650】【UOJ #219】【NOI 2016】优秀的拆分

http://www.lydsy.com/JudgeOnline/problem.php?id=4650
http://uoj.ac/problem/219
这里有非常好的题解qwq
接着道题复习一下模板（还是打错了好多地方_(:з」∠)_）
以及后缀数组面对多组数据清数组的正确打开方式qwq
因为求height数组时依赖n之后的字符都为空，倍增算法利用sa计算第一关键字时也依赖n之后的关键字都为空。
所以应至少在字符串第n+1位清空来终止可能会越界的求height数组的函数。还要把存第一关键字和第二关键字下标的数组都清空qwq
还是这种设关键点+差分的\(O(n\ln n)\)的方法也是一个经典模板。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 30013;

int t1[N], t2[N], c[N];

void st(int *x, int *y, int *sa, int n, int m) {
	for (int i = 0; i <= m; ++i) c[i] = 0;
	for (int i = 0; i < n; ++i) ++c[x[y[i]]];
	for (int i = 1; i <= m; ++i) c[i] += c[i - 1];
	for (int i = n - 1; i >= 0; --i) sa[--c[x[y[i]]]] = y[i];
}

void mkhz(int *r, int *sa, int n, int m) {
	memset(t1, 0, sizeof(t1));
	memset(t2, 0, sizeof(t2));
	int *x = t1, *y = t2, *t, i, j, p;
	for (i = 0; i < n; ++i) x[i] = r[i], y[i] = i;
	st(x, y, sa, n, m);
	for (j = 1, p = 1; j < n && p < n; j <<= 1, m = p - 1) {
		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 - 1]] == y[sa[i]] && y[sa[i - 1] + j] == y[sa[i] + j] ? p - 1 : p++;
	}
}

void mkh(int *r, int *sa, int *rank, int *h, int n) {
	h[0] = 0; int k = 0, j;
	for (int i = 0; i < n; ++i) rank[sa[i]] = i;
	for (int i = 1; i < n; h[rank[i++]] = k)
		for (k ? --k : k = 0, j = sa[rank[i] - 1]; r[i + k] == r[j + k]; ++k);
}

void pref(int *h, int f[N][16], int n) {
	for (int i = 1; i < n; ++i) f[i][0] = h[i];
	for (int j = 1; (1 << j) < n; ++j)
		for (int i = (1 << j); i < n; ++i)
			f[i][j] = min(f[i][j - 1], f[i - (1 << (j - 1))][j - 1]);
}

char s1[N], s2[N];
int r1[N], r2[N], f1[N][16], f2[N][16], sa1[N], sa2[N], h1[N], h2[N];
int Log_2[N], rank1[N], rank2[N], n, sum_pre[N], sum_suf[N];

int lcp(int i, int j) {
	i = rank1[i]; j = rank1[j];
	if (i > j) i ^= j ^= i ^= j;
	int l = Log_2[j - i];
	return min(f1[j][l], f1[i + (1 << l)][l]);
}

int lcs(int i, int j) {
	i = rank2[n - i + 1]; j = rank2[n - j + 1];
	if (i > j) i ^= j ^= i ^= j;
	int l = Log_2[j - i];
	return min(f2[j][l], f2[i + (1 << l)][l]);
}

int main() {
	int cnt = 0;
	for (int i = 1; i <= 30000; ++i) {
		if ((1 << (cnt + 1)) <= i) ++cnt;
		Log_2[i] = cnt;
	}
	int T; scanf("%d", &T);
	while (T--) {
		scanf("%s", s1 + 1);
		n = strlen(s1 + 1);
		for (int i = 1, j = n; i <= n; ++i, --j)
			r2[j] = r1[i] = s1[i] - 'a' + 1;
		r1[n + 1] = r2[n + 1] = 0;
		mkhz(r1, sa1, n + 1, 26);
		mkh(r1, sa1, rank1, h1, n + 1);
		mkhz(r2, sa2, n + 1, 26);
		mkh(r2, sa2, rank2, h2, n + 1);
		pref(h1, f1, n + 1);
		pref(h2, f2, n + 1);
		
		memset(sum_pre, 0, sizeof(int) * (n + 1));
		memset(sum_suf, 0, sizeof(int) * (n + 1));
		int x, y, l, r;
		for (int len = 1; (len << 1) <= n; ++len)
			for (int i = 1, j; (j = i + len) <= n; i = j) {
				x = lcs(i, j);
				y = lcp(i, j);
				if (x + y - 1 >= len) {
					++sum_pre[max(j, j + len - x)];
					--sum_pre[min(j + len, j + y)];
					++sum_suf[max(i - len + 1, i - x + 1)];
					--sum_suf[min(i + 1, i + y - len + 1)];
				}
			}
		
		for (int i = 1; i <= n; ++i) {
			sum_pre[i] += sum_pre[i - 1];
			sum_suf[i] += sum_suf[i - 1];
		}
		
		ll ans = 0;
		for (int i = 2; i <= n; ++i)
			ans += 1ll * sum_pre[i - 1] * sum_suf[i];
		printf("%lld\n", ans);
	}
	return 0;
}