DISUBSTR - Distinct Substrings

DISUBSTR - Distinct Substrings

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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.

分析：字符串中不同子串的个数；

　　　建立后缀数组对每一个后缀算贡献即可；

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define ld long double
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, ls[rt]
#define Rson mid+1, R, rs[rt]
#define sys system("pause")
#define freopen freopen("in.txt","r",stdin)
const int maxn=1e3+10;
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
inline ll read()
{
    ll x=0;int f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,k,t,cntA[maxn],cntB[maxn],sa[maxn],lev[maxn],height[maxn],A[maxn],B[maxn],tsa[maxn];
char ch[maxn];
void solve()
{
    for (int i = 0; i < 256; i ++) cntA[i] = 0;
    for (int i = 1; i <= n; i ++) cntA[ch[i]] ++;
    for (int i = 1; i < 256; i ++) cntA[i] += cntA[i - 1];
    for (int i = n; i; i --) sa[cntA[ch[i]] --] = i;
    lev[sa[1]] = 1;
    for (int i = 2; i <= n; i ++)
    {
        lev[sa[i]] = lev[sa[i - 1]];
        if (ch[sa[i]] != ch[sa[i - 1]]) lev[sa[i]] ++;
    }
    for (int l = 1; lev[sa[n]] < n; l <<= 1)
    {
        for (int i = 0; i <= n; i ++) cntA[i] = 0;
        for (int i = 0; i <= n; i ++) cntB[i] = 0;
        for (int i = 1; i <= n; i ++)
        {
            cntA[A[i] = lev[i]] ++;
            cntB[B[i] = (i + l <= n) ? lev[i + l] : 0] ++;
        }
        for (int i = 1; i <= n; i ++) cntB[i] += cntB[i - 1];
        for (int i = n; i; i --) tsa[cntB[B[i]] --] = i;
        for (int i = 1; i <= n; i ++) cntA[i] += cntA[i - 1];
        for (int i = n; i; i --) sa[cntA[A[tsa[i]]] --] = tsa[i];
        lev[sa[1]] = 1;
        for (int i = 2; i <= n; i ++)
        {
            lev[sa[i]] = lev[sa[i - 1]];
            if (A[sa[i]] != A[sa[i - 1]] || B[sa[i]] != B[sa[i - 1]]) lev[sa[i]] ++;
        }
    }
    for (int i = 1, j = 0; i <= n; i ++)
    {
        if (j) j --;
        while (ch[i + j] == ch[sa[lev[i] - 1] + j]) j ++;
        height[lev[i]] = j;
    }
}
int main()
{
    int i,j;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",ch+1);
        n=strlen(ch+1);
        solve();
        ll ans=0;
        rep(i,1,n)
        {
            ans+=n-sa[i]+1-height[i];
        }
        printf("%lld\n",ans);
    }
    //system("Pause");
    return 0;
}