spoj694 Distinct Substrings

SPOJ Problem Set (classical)

694. Distinct Substrings

Problem code: DISUBSTR

 

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.

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Added by:PrasannaDate:2006-01-13Time limit:1sSource limit:50000BLanguages:All except: PERL 6Resource:

ByteCode '06

 

 

 

 

 

 

 

 

 

#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int maxn=1010;char str[maxn];int wa[maxn],wb[maxn],wv[maxn],wn[maxn],a[maxn],sa[maxn];int cmp(int* r,int a,int b,int l){ return r[a]==r[b]&&r[a+l]==r[b+l];}//n为字符串长度，m为字符的取值范围，r为字符串。后面的j为每次排序时子串的长度void DA(int* r,int* sa,int n,int m){ int i,j,p,*x=wa,*y=wb,*t; ///对R中长度为1的子串进行基数排序 for(i=0; i<m; i++)wn[i]=0; for(i=0; i<n; i++)wn[x[i]=r[i]]++; for(i=1; i<m; i++)wn[i]+=wn[i-1]; for(i=n-1; i>=0; i--)sa[--wn[x[i]]]=i; for(j=1,p=1; p<n; j*=2,m=p) { //利用了上一次基数排序的结果，对待排序的子串的第二关键字进行了一次高效地基数排序 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; ///基数排序 for(i=0; i<n; i++)wv[i]=x[y[i]]; for(i=0; i<m; i++)wn[i]=0; for(i=0; i<n; i++)wn[wv[i]]++; for(i=1; i<m; i++)wn[i]+=wn[i-1]; for(i=n-1; i>=0; i--)sa[--wn[wv[i]]]=y[i]; ///当p=n的时候，说明所有串都已经排好序了 ///在第一次排序以后，rank数组中的最大值小于p，所以让m=p 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;}///后缀数组 计算height数组/**height数组的值应该是从height[1]开始的，而且height[1]应该是等于0的。原因是，+因为我们在字符串后面添加了一个0号字符，所以它必然是最小的一个后缀。而字符串中的其他字符都应该是大于0的（前面有提到，使用倍增算法前需要确保这点），所以排名第二的字符串和0号字符的公共前缀（即height[1]）应当为0.在调用calheight函数时，要注意height数组的范围应该是[1..n]。所以调用时应该是calheight(r,sa,n)而不是calheight(r,sa,n+1)。*/int rank[maxn],height[maxn];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 t; scanf("%d",&t); while(t--) { char str[1010]; scanf("%s",str); int n=strlen(str); for(int i=0;i<n;i++) a[i]=(int)str[i]; a[n]=0; DA(a,sa,n+1,260); calheight(a,sa,n); int sum=0; for(int i=1;i<=n;i++) sum+=n-sa[i]-height[i]; printf("%d/n",sum); } return 0;}