SP694 DISUBSTR - Distinct Substrings

题意翻译

给定一个字符串，求该字符串含有的本质不同的子串数量。

题目描述

Given a string, we need to find the total number of its distinct substrings.

输入输出格式

输入格式：

 

T- number of test cases. T<=20;
Each test case consists of one string, whose length is <= 1000

 

输出格式：

 

For each test case output one number saying the number of distinct substrings.

 

输入输出样例

输入样例#1： 复制

\n
  2\n
  CCCCC\n
  ABABA

输出样例#1： 复制

\n
  5\n
  9


Explanation for the testcase with string ABABA: \n
  len=1 : A,B\n
  len=2 : AB,BA\n
  len=3 : ABA,BAB\n
  len=4 : ABAB,BABA\n
  len=5 : ABABA\n

  Thus, total number of distinct substrings is 9.\n

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;

const int N=1e3+5;

char s[N];
int n,m,ans;
int rnk[N],sa[N],tax[N],tp[N];

inline void Qsort()
{
    for(int i=1;i<=m;++i)
        tax[i]=0;
    for(int i=1;i<=n;++i)
        ++tax[rnk[i]];
    for(int i=1;i<=m;++i)
        tax[i]+=tax[i-1];
    for(int i=n;i;--i)
        sa[tax[rnk[tp[i]]]--]=tp[i];
}

inline void Suffix_Sort()
{
    for(int i=1;i<=n;++i)
        rnk[i]=s[i],tp[i]=i;
    Qsort();
    for(int l=1,p=0;p<n;m=p,l<<=1)
    {
        p=0;
        for(int i=n-l+1;i<=n;++i)
            tp[++p]=i;
        for(int i=1;i<=n;++i)
            if(sa[i]>l)
                tp[++p]=sa[i]-l;
        Qsort();
        for(int i=1;i<=n;++i)
            swap(rnk[i],tp[i]);
        rnk[sa[1]]=p=1;
        for(int i=2;i<=n;++i)
            rnk[sa[i]]=(tp[sa[i]]==tp[sa[i-1]]&&tp[sa[i]+l]==tp[sa[i-1]+l])?p:++p;
    }
}

inline void Get_Height()
{
    int j,k=0;
    for(int i=1;i<=n;++i)
    {
        if(k)
            --k;
        j=sa[rnk[i]-1];
        while(j+k<=n&&i+k<=n&&s[i+k]==s[j+k])
            ++k;
        ans-=k;
    }
}

int T;
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s",s+1);
        n=strlen(s+1),m=N-1;
        ans=n*(n+1)/2;
        Suffix_Sort();
        Get_Height();
        cout<<ans<<'\n';
    }
    return 0;
}
/*
2
CCCCC
ABABA
*/