SPOJ Distanct Substrings(求不同子串的数量)

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.

题解：题意就是让你求子串的种类；

思路：后缀数组，然后每个字符的贡献为n-sa[i]-height[i];(n为字符的个数)；

参考代码：

 //#include<bits/stdc++.h>
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<string>
 #include<cstdlib>
 #include<cmath>
 #include<algorithm>
 using namespace std;
 #define clr(a,val) memset(a,val,sizeof(a))
 #define lowbit(x) x&-x
 #define eps 1e-6
 typedef long long ll;
 const int INF=0x3f3f3f3f;
 const int maxn=;
 struct SuffixArray{
     int s[maxn];
     int sa[maxn],height[maxn],rank[maxn],n;
     int t[maxn*],t2[maxn*];
     int cnt[maxn];
     void build_sa(int m)//字符都属于0~m-1范围
     {
         int i,*x=t,*y=t2;
         for(i=;i<m;i++) cnt[i]=;
         for(i=;i<n;i++) cnt[x[i]=s[i]]++;
         for(i=;i<m;i++) cnt[i]+=cnt[i-];
         for(i=n-;i>=;i--) sa[--cnt[x[i]]]=i;
         for(int k=,p;k<=n;k <<=)//k<=n
         {
             p=;
             for(i=n-k;i<n;i++) y[p++]=i;
             for(i=;i<n;i++) if(sa[i]>=k) y[p++]=sa[i]-k;
             for(i=;i<m;i++) cnt[i]=;
             for(i=;i<n;i++) cnt[x[y[i]]]++;
             for(i=;i<m;i++) cnt[i]+=cnt[i-];
             for(i=n-;i>=;i--) sa[--cnt[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-]+k]==y[sa[i]+k]? p-:p++;
             if(p>=n) break;
             m=p;
         }
     }
     void build_height()
     {
         int k=;
         for(int i=;i<n;i++) rank[sa[i]]=i;
         for(int i=;i<n-;i++)
         {
             if(k) k--;
             int j=sa[rank[i]-];
             while(s[i+k]==s[j+k]) k++;
             height[rank[i]]=k;
         }
     }
 } SA;
 int main()
 {
     int n, m, t;
     string str;
     cin>>t;
     while(t--)
     {
         cin>>str;
         int n = str.size();
         for(int i = ; i<n; i++) SA.s[i] = str[i];
         SA.s[n] = ;SA.n=n+;
         SA.build_sa();
         SA.build_height();
         int ans = ;
         for(int i=;i<=n;i++) ans+=n-SA.sa[i]-SA.height[i];
         cout<<ans<<endl;
     }
     return ;
 }