SPOJ Distinct Substrings【后缀数组】

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：

求一个字符串中所有的不相同的子串个数。

思路：

一个字符串的所有子串个数是n*(n+1)/2，关键就是有多少是重复的。

一个子串其实就是原字符串某个后缀的前缀，找到有多少重复的其实就是找后缀的lcp

比如我们遍历到了第k个后缀，本来他可以产生len个前缀，但是他的前面有一部分前缀是已经算过的了。

而且由于 LCP(i,j)=min{LCP(k-1,k)|i+1≤k≤j} ，且对 i≤j<k，LCP(j,k)≥LCP(i,k)

每加入一个后缀，应该减去max(LCP(i,j)), 也就是LCP（i-1,i），也就是height[i]

 #include <iostream>

 #include <set>

 #include <cmath>

 #include <stdio.h>

 #include <cstring>

 #include <algorithm>

 #include <vector>

 #include <queue>

 #include <map>

 using namespace std;

 typedef long long LL;

 #define inf 0x7f7f7f7f

 const int maxn = ;

 int t;

 char str[maxn];

 int sa[maxn];

 int t1[maxn], t2[maxn], c[maxn];

 int rnk[maxn], height[maxn];

 void build_sa(int s[], int n, int m)

 {

     int i, j, p, *x = t1, *y = t2;

     for(i = ; i < m; i++)c[i] = ;

     for(i = ; i < n; i++)c[x[i] = s[i]]++;

     for(i = ; i < m; i++)c[i] += c[i - ];

     for(i = n - ; i >= ; i--)sa[--c[x[i]]] = i;

     for(j = ; j <= n; j <<= ){

         p = ;

         for(i = n - j; i < n; i++)y[p++] = i;

         for(i = ; i < n; i++)if(sa[i] >= j)y[p++] = sa[i] - j;

         for(i = ; i < m; i++)c[i] = ;

         for(i = ; i < n; i++)c[x[y[i]]]++;

         for(i = ; i < m; i++)c[i] += c[i - ];

         for(i = n - ; i >= ; i--)sa[--c[x[y[i]]]] = y[i];

         swap(x, y);

         p = ;

         x[sa[]] = ;

         for(i = ; i < n; i++)

             x[sa[i]] = y[sa[i - ]] == y[sa[i]] && y[sa[i - ] + j] == y[sa[i] + j] ? p - :p++;

         if(p >= n)break;

         m = p;

     }

 }

 void get_height(int s[], int n)

 {

     int i, j, k = ;

     for(i = ; i <= n; i++)rnk[sa[i]] = i;

     for(i = ; i < n; i++){

         if(k) k--;

         j = sa[rnk[i] - ];

         while(s[i + k] == s[j + k])k++;

         height[rnk[i]] = k;

     }

 }

 int s[maxn];

 int main()

 {

     scanf("%d", &t);

     while(t--){

         scanf("%s", str);

         int n = strlen(str);

         for(int i = ; i <= n; i++)s[i] = str[i];

         build_sa(s, n + , );

         get_height(s, n);

         LL ans = n * (n + ) / ;

         //cout<<ans<<endl;

         for(int i = ; i <= n; i++){

             ans -= height[i];

         }

         printf("%lld\n", ans);

     }

     return ;

 }