SPOJ694 -- DISUBSTR 后缀树组求不相同的子串的个数

DISUBSTR - Distinct Substrings

 

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：求不同的子串的个数。

首先要知道，任意子串必是某一后缀的前缀。对于suffix[sa[i]],  它有len-sa[i]个前缀，，其中lcp[i]个子串与suffix[sa[i-1]]的前缀重复。。

每次只需要加上 len-sa[i]-lcp[i]就行了。。  

#include <set>
#include <map>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const int maxn = 2e4+10;
int sa[maxn], k, len, tmp[maxn], rank[maxn];
string s;
bool cmp(int i, int j)
{
    if (rank[i] != rank[j])
        return rank[i] < rank[j];
    else
    {
        int x = (i+k <= len ? rank[i+k] : -1);
        int y = (j+k <= len ? rank[j+k] : -1);
        return x < y;
    }
}
void build_sa()
{
    for (int i = 0; i <= len; i++)
    {
        sa[i] = i;
        rank[i] = (i < len ? s[i] : -1);
    }
    for (k = 1; k <= len; k *= 2)
    {
        sort (sa,sa+len+1,cmp);
        tmp[sa[0]] = 0;
        for (int i = 1; i <= len; i++)
        {
            tmp[sa[i]] = tmp[sa[i-1]] + (cmp(sa[i-1],sa[i])? 1 : 0);
        }
        for (int i = 0; i <= len; i++)
            rank[i] = tmp[i];
    }
}
int lcp[maxn];
void get_lcp()
{
    for (int i = 0; i < len; i++)
        rank[sa[i]] = i;
    int h = 0;
    lcp[0] = 0;
    for (int i = 0; i < len; i++)
    {
        int j = sa[rank[i]-1];
        if (h > 0)
            h--;
        for (; h+i < len && h+j < len; h++)
            if (s[i+h] != s[j+h])
                break;
        lcp[rank[i]] = h;
    }
}
int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
    #endif
    int T;
    scanf ("%d", &T);
    while (T--)
    {
        cin >> s;
        len = s.size();
        build_sa();
        get_lcp();
        int ans = 0;
        for (int i = 0; i <= len; i++)
        {
            ans += len - sa[i] - lcp[i];
        }
        printf ("%d
",ans);
    }
    return 0;
}