SPOJ New Distinct Substrings

D - New Distinct Substrings

Time Limit:280MS     Memory Limit:1572864KB     64bit IO Format:%lld & %llu

Submit  Status

Description

Given a string, we need to find the total number of its distinct substrings.

Input

T- number of test cases. T<=20; Each test case consists of one string, whose length is <= 50000

Output

For each test case output one number saying the number of distinct substrings.

Example

Input:
2
CCCCC
ABABA

Output:
5
9

#include <cstdio>
#include <cstring>

const int N = 50010;
char str[N];
int sa[N], wa[N], wb[N], ws[N], wv[N];

bool Cmp(int *r, int a, int b, int l) {
	return r[a] == r[b] && r[a + l] == r[b + l];
}

void DA(int n, int m) {
	int i, j, p, *x = wa, *y = wb, *t;
	for (i = 0; i < m; ++i) ws[i] = 0;
	for (i = 0; i < n; ++i) ws[x[i] = str[i]]++;
	for (i = 1; i < m; ++i) ws[i] += ws[i - 1];
	for (i = n - 1; i >= 0; --i) sa[--ws[x[i]]] = i;

	for (j = p = 1; p < n; m = p, j <<= 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;

		for (i = 0; i < n; ++i) wv[i] = x[y[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) sa[--ws[wv[i]]] = y[i];

		t = x, x = y, y = t;
		for (p = i = 1, x[sa[0]] = 0; i < n; ++i)
			x[sa[i]] = Cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++;
	}
}

int height[N], rank[N];

void calheight(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]; str[i + k] == str[j + k]; ++k);
}

int main() {
	int T;
	scanf("%d", &T);

	while (T--) {
		int ans = 0;
		scanf("%s", str);
		int len = strlen(str);

		DA(len + 1, 256);
		calheight(len);

		for (int i = 1; i <= len; ++i)
			ans += (len - sa[i] - height[i]);

		printf("%d\n", ans);
	}

	return 0;
}