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=1010;
struct SuffixArray{
    int s[maxn];
    int sa[maxn],height[maxn],rank[maxn],n;
    int t[maxn*2],t2[maxn*2];
    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=0;i<m;i++) cnt[i]=0;
        for(i=0;i<n;i++) cnt[x[i]=s[i]]++;
        for(i=1;i<m;i++) cnt[i]+=cnt[i-1];
        for(i=n-1;i>=0;i--) sa[--cnt[x[i]]]=i; 
        for(int k=1,p;k<=n;k <<=1)//k<=n
        {
            p=0;
            for(i=n-k;i<n;i++) y[p++]=i;
            for(i=0;i<n;i++) if(sa[i]>=k) y[p++]=sa[i]-k;
            for(i=0;i<m;i++) cnt[i]=0;
            for(i=0;i<n;i++) cnt[x[y[i]]]++;
            for(i=1;i<m;i++) cnt[i]+=cnt[i-1];
            for(i=n-1;i>=0;i--) sa[--cnt[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]+k]==y[sa[i]+k]? p-1:p++;
            if(p>=n) break;
            m=p;
        }
    }
    void build_height()
    {
        int k=0;
        for(int i=0;i<n;i++) rank[sa[i]]=i;
        for(int i=0;i<n-1;i++)
        {
            if(k) k--;
            int j=sa[rank[i]-1];
            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 = 0; i<n; i++) SA.s[i] = str[i];
        SA.s[n] = 0;SA.n=n+1;
        SA.build_sa(128);
        SA.build_height();
        int ans = 0;
        for(int i=1;i<=n;i++) ans+=n-SA.sa[i]-SA.height[i];
        cout<<ans<<endl;
    }
    return 0;
}