SPOJ SUBST1 New Distinct Substrings

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SUBST1 - New 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 <= 50000

Output

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

Example

Input:2CCCCCABABAOutput:59

后缀数组模板题

#include <stdio.h>#include <string.h>const int MAXN=50005;int wa[MAXN],wb[MAXN],wv[MAXN],ws[MAXN],rank[MAXN],height[MAXN], r[MAXN],sa[MAXN];char s[MAXN];int cmp(int *r,int a,int b,int l){        if(r[a]==r[b]&&r[a+l]==r[b+l])                return 1;        return 0;}void da(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++;        }        return;}void calheight(int *r,int *sa,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++);        return;}int main(){        int i,n,t;        long long ans;        scanf("%d",&t);        while(t-->0)        {                scanf("%s",s);                n=strlen(s);                for(i=0;i<n;i++)                        r[i]=s[i];                r[n]=0;                da(r,sa,n+1,128);                calheight(r,sa,n);                ans=(long long)n*(n+1)/2;                for(i=1;i<=n;i++)                        ans-=height[i];                printf("%lld\n",ans);        }    return 0;}

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