SPOJ DISUBSTR - Distinct Substrings or SUBST1 - New Distinct Substrings 【不同子串数目】

题目1链接：SPOJ DISUBSTR - Distinct Substrings
题目1链接：SPOJ SUBST1 - New Distinct Substrings

DISUBSTR - Distinct Substrings
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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.

Example

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

题意：给定一个字符串，求不同子串的数目。

思路：每一个子串一定是某些后缀的前缀，我们只需要统计所有后缀里面的不同前缀即可。对于后缀sa[i]而言，它的前缀有n-sa[i]个。而对于两个相邻的后缀sa[i-1]，sa[i]，重复的前缀有H[i]个，我们从前到后扫一遍即可。

第二道题把数组开大点就OK了。

AC代码：

//#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <algorithm>#include <vector>#include <queue>#include <map>#include <stack>#define PI acos(-1.0)#define CLR(a, b) memset(a, (b), sizeof(a))#define fi first#define se second#define ll o<<1#define rr o<<1|1using namespace std;typedef long long LL;typedef pair<int, int> pii;const int MAXN = 2*1e3 + 10;const int pN = 1e6;// <= 10^7const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;void add(LL &x, LL y) { x += y; x %= MOD; }int cmp(int *r, int a, int b, int l) {    return (r[a] == r[b]) && (r[a+l] == r[b+l]);}int wa[MAXN], wb[MAXN], ws[MAXN], wv[MAXN];int R[MAXN];//下标0->n-1 存储的是1->n之间的数int H[MAXN];//下标2->n   H[i] >= H[i-1]-1void DA(int *r, int *sa, 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]=r[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 = 1, p = 1; p < n; j *= 2, m = p)    {        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];        for(t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; i++)            x[sa[i]] = cmp(y, sa[i-1], sa[i], j) ? p-1 : p++;    }}void calh(int *r, int *sa, int n){    int i, j, k = 0;    for(i = 1; i <= n; i++) R[sa[i]] = i;    for(i = 0; i < n; H[R[i++]] = k)    for(k ? k-- : 0, j = sa[R[i]-1]; r[i+k] == r[j+k]; k++);}int sa[MAXN];//下标从1->n，存储的是0->n-1之间的数int a[MAXN];char str[MAXN];int main(){    int t; scanf("%d", &t);    while(t--) {        scanf("%s", str);        int n = strlen(str); int Max = 0;        for(int i = 0; i < n; i++) {            a[i] = str[i];            Max = max(Max, a[i]);        }        a[n] = 0;        DA(a, sa, n+1, Max+1);//<n+1        calh(a, sa, n);//<=n        int ans = 0;        for(int i = 1; i <= n; i++) {            if(i == 1) ans += n - sa[i];            else {                ans += n - sa[i] - H[i];            }        }        printf("%d\n", ans);    }    return 0;}