SPOJ- Distinct Substrings(后缀数组&后缀自动机)

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：

本题题意就是给你一个字符串，让你找它有多少不同的子串；

其实就是SAM的板题，只要求每一个状态点的longest[i]-longest[fa[i]]的和就行了。

但由于是后缀数组专题，还是用后缀数组写：

参考代码：

后缀自动机：

#include<bits/stdc++.h>
using namespace std;
#define PI acos(-1.0)
#define mkp make_pair
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
typedef long long ll;
const int INF=0x3f3f3f3f;
const int maxn=1e5+;
char s[maxn];
struct SAM{
    ll ans;
    int fa[maxn<<],l[maxn<<],nxt[maxn<<][],last,cnt;
    void Init()
    {
        memset(nxt[],,sizeof(nxt[]));
        last=cnt=; ans=;
        fa[]=;l[]=;
    }

    int NewNode()
    {
        ++cnt;
        memset(nxt[cnt],,sizeof(nxt[cnt]));
        fa[cnt]=l[cnt]=;
        return cnt;
    }

    void Add(int c)
    {
        int p=last,np=NewNode();
        last=np;l[np]=l[p]+;
        while(p&&!nxt[p][c]) nxt[p][c]=np,p=fa[p];
        if(!p) fa[np]=;
        else
        {
            int q=nxt[p][c];
            if(l[q]==l[p]+) fa[np]=q;
            else
            {
                int nq=NewNode();
                memcpy(nxt[nq],nxt[q],sizeof(nxt[q]));
                fa[nq]=fa[q];
                l[nq]=l[p]+;
                fa[q]=fa[np]=nq;
                while(nxt[p][c]==q) nxt[p][c]=nq,p=fa[p];
            }
        }
        ans+=(l[last]-l[fa[last]])*1ll;
    }

    void Query()
    {
        Init();
        for(int i=,len=strlen(s);i<len;++i) Add(s[i]-'A');
        printf("%lld\n",ans);
    }
} sam;

int main()
{
    int N;
    scanf("%d",&N);
    while(N--)
    {
        sam.Init();
        scanf("%s",s);
        sam.Query();
    }
    return ;
}

后缀数组：

#include<iostream>
#include<cstdio>
#include<cstring>
#define rint register int
#define ini inline int
#define maxn 1000050
using namespace std;
char str[maxn];
int y[maxn<<],x[maxn<<],c[maxn];
int sa[maxn],rk[maxn],height[maxn];
int n,m,s[maxn];

inline void get_SA()
{
    for(int i=;i<=m;++i) c[i]=;
    for(int i=;i<=n;++i) ++c[x[i]=s[i]];
    for(int i=;i<=m;++i) c[i]+=c[i-];
    for(int i=n;i>=;--i) sa[c[x[i]]--]=i;
    for(int k=;k<=n;k<<=)
    {
        int num=;
        for(int i=n-k+;i<=n;++i) y[++num]=i;
        for(int i=;i<=n;++i) if(sa[i]>k) y[++num]=sa[i]-k;
        for(int i=;i<=m;++i) c[i]=;
        for(int i=;i<=n;++i) ++c[x[i]];
        for(int i=;i<=m;++i) c[i]+=c[i-];
        for(int i=n;i>=;--i) sa[c[x[y[i]]]--]=y[i],y[i]=;
        swap(x,y);
        x[sa[]]=;
        num=;
        for(rint i=;i<=n;++i)
            x[sa[i]]=(y[sa[i]]==y[sa[i-]]&&y[sa[i]+k]==y[sa[i-]+k])?num:++num;
        if(num==n) break;
        m=num;
    }
}
inline void get_height()
{
    int k=;
    for(int i=;i<=n;++i) rk[sa[i]]=i;
    for(int i=;i<=n;++i)
    {
        if(rk[i]==) continue;//第一名height为0
        if(k) --k;//h[i]>=h[i-1]-1;
        rint j=sa[rk[i]-];
        while(j+k<=n&&i+k<=n&&s[i+k]==s[j+k]) ++k;
        height[rk[i]]=k;//h[i]=height[rk[i]];
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s",str+);
        n=strlen(str+);m=;
        for(int i=;i<=n;++i) s[i]=str[i]-'A'+;
        get_SA();
        get_height();
        //for(int i=1;i<=n;++i) cout<<sa[i]<<" "<<height[i]<<endl;
        int ans=;
        for(int i=;i<=n;++i) ans+=n-sa[i]+-height[i];
        printf("%d\n",ans);
    }
    return ;
}