SPOJ

Distinct Substrings 

Time Limit: 159MS

Memory Limit: 1572864KB

64bit IO Format: %lld & %llu

Description

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.

/*
 * SPOJ - DISUBSTR  Distinct Substrings
 * 求不同子串的个数
 *
 * 长度为n的字符串有n*(n+1)/2个子串，再减去相同的子串就行了
 * 对于子串，它肯定是一个后缀的前缀,如果height[i]==k,说明后缀i-1和后缀i有k个子串相同，
 * 这样减去它即可，即减去height数组的后n-1个即可
 */

#include <bits/stdc++.h>
using namespace std;

const int MAXN = 1000+100;

int sa[MAXN];
int t1[MAXN],t2[MAXN],c[MAXN];
int Rank[MAXN],height[MAXN];
void build_sa(int s[],int n,int m)
{
    int i,j,p,*x=t1,*y=t2;
    for(i=0;i<m;i++)c[i]=0;
    for(i=0;i<n;i++)c[x[i]=s[i]]++;
    for(i=1;i<m;i++)c[i]+=c[i-1];
    for(i=n-1;i>=0;i--)sa[--c[x[i]]]=i;
    for(j=1;j<=n;j<<=1)
    {
        p=0;
        for(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<m;i++)c[i]=0;
        for(i=0;i<n;i++)c[x[y[i]]]++;
        for(i=1;i<m;i++)c[i]+=c[i-1];
        for(i=n-1;i>=0;i--)sa[--c[x[y[i]]]]=y[i];
        swap(x,y);
        p=1;x[sa[0]]=0;
        for(i=1;i<n;i++)
            x[sa[i]]=y[sa[i-1]]==y[sa[i]] && y[sa[i-1]+j]==y[sa[i]+j]?p-1:p++;
        if(p>=n)break;
        m=p;
    }
}
void getHeight(int s[],int n)
{
    int i,j,k=0;
    for(i=0;i<=n;i++) Rank[sa[i]]=i;
    for(i=0;i<n;i++)
    {
        if(k)k--;
        j=sa[Rank[i]-1];
        while(s[i+k]==s[j+k])k++;
        height[Rank[i]]=k;
    }
}
char str[MAXN];
int ss[MAXN];

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s",str);
        int len=strlen(str);
        for(int i=0;i<len;i++) ss[i]=str[i];
        ss[len]=0;
        build_sa(ss,len+1,128);
        getHeight(ss,len);
        int ans=len*(len+1)/2;
        for(int i=2;i<=len;i++) ans-=height[i];
        printf("%d
",ans);
    }
    return 0;
}