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SPOJ 687 Repeats 后缀数组

和上一题差不多的方法。。没什么好说的

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = (5e4 + 10) * 4;#define F(x) ((x) / 3 + ((x) % 3 == 1 ? 0 : tb))#define G(x) ((x) < tb ? (x) * 3 + 1 : ((x) - tb) * 3 + 2)int wa[maxn], wb[maxn], wv[maxn], ws[maxn];int c0(int *r, int a, int b) {	return r[a] == r[b] && r[a + 1] == r[b + 1] && r[a + 2] == r[b + 2];}int c12(int k, int *r, int a, int b) {	if (k == 2) return r[a] < r[b] || r[a] == r[b] && c12(1, r, a + 1, b + 1);	else return r[a] < r[b] || r[a] == r[b] && wv[a + 1] < wv[b + 1];}void sort(int *r, int *a, int *b, int n, int m) {	int i;	for (i = 0; i < n; i++) wv[i] = r[a[i]];	for (i = 0; i < m; i++) ws[i] = 0;	for (i = 0; i < n; i++) ws[wv[i]]++;	for (i = 1; i < m; i++) ws[i] += ws[i - 1];	for (i = n - 1; i >= 0; i--) b[--ws[wv[i]]] = a[i];}void dc3(int *r, int *sa, int n, int m) {	int i, j, *rn = r + n, *san = sa + n, ta = 0, tb = (n + 1) / 3, tbc = 0, p;	r[n] = r[n + 1] = 0;	for (i = 0; i < n; i++) if (i % 3 != 0) wa[tbc++] = i;	sort(r + 2, wa, wb, tbc, m);	sort(r + 1, wb, wa, tbc, m);	sort(r, wa, wb, tbc, m);	for (p = 1, rn[F(wb[0])] = 0, i = 1; i < tbc; i++)		rn[F(wb[i])] = c0(r, wb[i - 1], wb[i]) ? p - 1 : p++;	if (p < tbc) dc3(rn, san, tbc, p);	else for (i = 0; i < tbc; i++) san[rn[i]] = i;	for (i = 0; i < tbc; i++) if (san[i] < tb) wb[ta++] = san[i] * 3;	if (n % 3 == 1) wb[ta++] = n - 1;	sort(r, wb, wa, ta, m);	for (i = 0; i < tbc; i++) wv[wb[i] = G(san[i])] = i;	for (i = 0, j = 0, p = 0; i < ta && j < tbc; p++)		sa[p] = c12(wb[j] % 3, r, wa[i], wb[j]) ? wa[i++] : wb[j++];	for (; i < ta; p++) sa[p] = wa[i++];	for (; j < tbc; p++) sa[p] = wb[j++];}int Rank[maxn], height[maxn];void calheight(int *r, int *sa, int n) {	int i, j, k = 0;	for (i = 1; i <= n; i++) Rank[sa[i]] = i;	for (i = 0; i < n; height[Rank[i++]] = k)	for (k ? k-- : 0, j = sa[Rank[i] - 1]; r[i + k] == r[j + k]; k++);}#undef F#undef Gint str[maxn], sa[maxn], T, n, minv[maxn][30];void init_RMQ() {	for(int i = 0; i <= n; i++) {		minv[i][0] = height[i];	}	for(int j = 1; (1 << j) <= n + 1; j++) {		for(int i = 0; i + (1 << j) - 1 <= n; i++) {			minv[i][j] = min(minv[i][j - 1], minv[i + (1 << (j - 1))][j - 1]);		}	}}int query(int ql, int qr) {	if(ql > qr) swap(ql, qr);	ql++;	int k = 0;	while((1 << (k + 1)) <= qr - ql + 1) k++;	return min(minv[ql][k], minv[qr - (1 << k) + 1][k]);}int main() {	scanf("%d", &T);	while(T--) {		scanf("%d", &n);		for(int i = 0; i < n; i++) {			char tmp; scanf(" %c", &tmp);			str[i] = tmp;		}		str[n] = 0;		dc3(str, sa, n + 1, 200);		calheight(str, sa, n);		init_RMQ();		int maxstep = 1;		for(int L = 1; L <= n; L++) {			for(int i = 0; i + L < n; i += L)  {				int s1 = query(Rank[i], Rank[i + L]);				int step = s1 / L + 1, ql = i - (L - s1 % L);				if(ql >= 0) {					step = max(step, query(Rank[ql], Rank[ql + L]) / L + 1);				}				maxstep = max(maxstep, step);			}		}		printf("%d\n", maxstep);	}	return 0;}

 

SPOJ 687 Repeats 后缀数组