SPOJ 694 Distinct Substrings

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

【题目分析】

求不同的子串个数，对于一个字串sa[i]，它有height[i]个字串是和前面一样的，删去即可，只需要扫一遍就可以得到答案了

【代码】

#include <cstring>
#include <cstdio>
#define maxn 20001
#define m(a) memset(a,0,sizeof a)
int len,wa[maxn],wb[maxn],wv[maxn],ws[maxn],sa[maxn],rank[maxn],height[maxn],s[maxn],n,a[maxn];
char ch[maxn];
inline int cmp(int *r,int a,int b,int l)
{return r[a]==r[b]&&r[a+l]==r[b+l];}
inline void getsa(int *r,int *sa,int n,int m)
{
int i,j,p,*x=wa,*y=wb,*t;
for (i=0;i<m;++i) ws[i]=0;
for (i=0;i<n;++i) ws[x[i]=r[i]]++;
for (i=1;i<m;++i) ws[i]+=ws[i-1];
for (i=n-1;i>=0;--i) sa[--ws[x[i]]]=i;
for (j=1,p=1;p<n;j*=2,m=p)
{
for (p=0,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<n;++i) wv[i]=x[y[i]];
for (i=0;i<m;++i) ws[i]=0;
for (i=0;i<n;++i) ws[wv[i]]++;
for (i=1;i<m;++i) ws[i]+=ws[i-1];
for (i=n-1;i>=0;--i) sa[--ws[wv[i]]]=y[i];
for (t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1;i<n;++i)
x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;
}
}
inline void gethi(int *r,int n)
{
int i,j,k=0;
for (i=1;i<=n;++i) rank[sa[i]]=i;
for (i=0;i<n;height[rank[i++]]=k)
for (k?k--:0,j=sa[rank[i]-1];r[i+k]==r[j+k];k++);
}
inline bool test(int k)
{
int i=2;
while (i<=n)
{
if (height[i]>=k)
{
int tot=2;
while (height[i+1]>=k) ++i,++tot;
if (tot>=len) return true;
}
i++;
}
return false;
}
int main()
{
int tt;
scanf("%d",&tt);
for (int j=1;j<=tt;++j)
{
scanf("%s",ch);
int l=strlen(ch);
for (int i=0;i<l;++i) s[i]=(int)ch[i];
s[l]=0;
getsa(s,sa,l+1,200);
gethi(s,l);
int ans=0;
ans+=l-sa[1];
for (int i=2;i<=l;++i)
ans+=l-sa[i]-height[i];
printf("%d\n",ans);
}
}