spoj694 Distinct Substrings 后缀数组

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:50000BMemory limit:256MBCluster:Pyramid (Intel Pentium III 733 MHz)Languages:All except: NODEJS PERL 6Resource:ByteCode '06

题意：给定一个字符串，求不相同的子串的个数。

思路： 每个子串一定是某个后缀的前缀，那么原问题等价于求所有后缀之前的不相同的前缀的个数。如果所有的后缀按照suffix(sa[1]),suffix(sa[2]),...suffix(sa[n])的顺序计算，不难发现，对于每次新加的后缀suffix(sa[k]),若n为不包含0的字符串的长度，则产生n-sa[k]个新的前缀，但其中height[k]和前面的字符串的前缀是相同的，锁以suffix(sa[k])将贡献n-sa[k]-height[k]个不同的子串，累加后是原问题的答案。时间复杂度为O(n)。详见代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int MAXN=1000+100;int n;char s[MAXN];int t1[MAXN],t2[MAXN],c[MAXN],rank[MAXN],height[MAXN],sa[MAXN];int cmp(int *r,int a,int b,int l){    return r[a]==r[b] && r[a+l]==r[b+l];}void build_sa(int m){    int i,k,p=0;    int *x=t1,*y=t2;    for(i=0;i<m;i++) c[i]=0;    for(i=0;i<n;i++) c[x[i]=s[i]]++;    for(i=1;i<m;i++) c[i]+=c[i-1];    for(i=n-1;i>=0;i--) sa[--c[x[i]]]=i;    for(k=1;p<n;k<<=1,m=p)    {        for(p=0,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++) c[i]=0;        for(i=0;i<n;i++) c[x[y[i]]]++;        for(i=1;i<m;i++) c[i]+=c[i-1];        for(i=n-1;i>=0;i--) sa[--c[x[y[i]]]]=y[i];        swap(x,y);        for(p=1,x[sa[0]]=0,i=1;i<n;i++)            x[sa[i]]=cmp(y,sa[i-1],sa[i],k)? p-1:p++;    }}void calheight(int n){    int k=0;    for(int i=1;i<=n;i++) rank[sa[i]]=i;    for(int i=0;i<n;i++)    {        if(k) k--;        int j=sa[rank[i]-1];        while(s[i+k]==s[j+k]) k++;        height[rank[i]]=k;    }}int main(){    freopen("text.txt","r",stdin);    int T;    scanf("%d",&T);    while(T--)    {        scanf("%s",s);        n=strlen(s);        s[n++]=0;        build_sa(200);        calheight(n-1);        int len=n-1;        int ans=len-sa[1];        for(int i=2;i<n;i++)            ans+=len-sa[i]-height[i];        printf("%d\n",ans);    }    return 0;}