1. 程式人生 > >SPOJ Distinct Substrings (字尾陣列,不相同的子串個數)

SPOJ Distinct Substrings (字尾陣列,不相同的子串個數)

描述

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 <= 1000

Output

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

Sample Input:

2
CCCCC
ABABA

Sample Output:

5
9

Explanation for the testcase with string ABABA:
len=1 : A,B
len=2 : AB,BA
len=3 : ABA,BAB
len=4 : ABAB,BABA
len=5 : ABABA
Thus, total number of distinct substrings is 9.

思路

給出一個字串,求其中不同的子串個數.

一個串的子串個數為n*(n+1)/2,減去後綴的相同字首所作的貢獻每一個height[i]

即為答案.

引用論文:

TIM截圖20180824164632.png

程式碼

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
#define mem(a, b) memset(a, b, sizeof(a))
const int N = 1e5 + 10;
int t1[N], t2[N], c[N], r[N], sa[N], rak[N], height[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 str[], int sa[], int rak[], int height[], int n, int m) { n++; int i, j, p, *x = t1, *y = t2; for (i = 0; i < m; i++) c[i] = 0; for (i = 0; i < n; i++) c[x[i] = str[i]]++; for (i = 1; i < m; i++) c[i] += c[i - 1]; for (i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i; for (j = 1; j <= n; j <<= 1) { p = 0; for (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 < m; i++) c[i] = 0; for (i = 0; i < n; i++) c[x[y[i]]]++; for (i = 1; i < m; i++) c[i] += c[i - 1]; for (i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i]; swap(x, y); p = 1; x[sa[0]] = 0; for (i = 1; i < n; i++) x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++; if (p >= n) break; m = p; } int k = 0; n--; for (i = 0; i <= n; i++) rak[sa[i]] = i; for (i = 0; i < n; i++) { if (k) k--; j = sa[rak[i] - 1]; while (str[i + k] == str[j + k]) k++; height[rak[i]] = k; } } char s[N]; int a[N]; int main() { //freopen("in.txt", "r", stdin); int t; scanf("%d", &t); while (t--) { scanf("%s", s); int n = strlen(s); for (int i = 0; i < n; i++) a[i] = s[i]; a[n] = 0; da(a, sa, rak, height, n, 128); int ans = n * (n + 1) / 2; for (int i = 2; i <= n; i++) ans -= height[i]; printf("%d\n", ans); } return 0; }