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 ;
 }