SPOJ 694（后缀数组）
694. Distinct SubstringsProblem 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.
#include <stdio.h> #include <string.h> const int maxn = 10005*2; #define F(x) ((x)/3+((x)%3==1?0:tb)) #define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2) int wa[maxn],wb[maxn],wv[maxn],ws[maxn],a[maxn],sa[maxn]; char str1[maxn]; int len1,sum; int c0(int *r,int a,int b) {return r[a]==r[b]&&r[a+1]==r[b+1]&&r[a+2]==r[b+2];} int c12(int k,int *r,int a,int b) {if(k==2) return r[a]<r[b]||r[a]==r[b]&&c12(1,r,a+1,b+1); else return r[a]<r[b]||r[a]==r[b]&&wv[a+1]<wv[b+1];} void sort(int *r,int *a,int *b,int n,int m) { int i; for(i=0;i<n;i++) wv[i]=r[a[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--) b[--ws[wv[i]]]=a[i]; return; } void dc3(int *r,int *sa,int n,int m) { int i,j,*rn=r+n,*san=sa+n,ta=0,tb=(n+1)/3,tbc=0,p; r[n]=r[n+1]=0; for(i=0;i<n;i++) if(i%3!=0) wa[tbc++]=i; sort(r+2,wa,wb,tbc,m); sort(r+1,wb,wa,tbc,m); sort(r,wa,wb,tbc,m); for(p=1,rn[F(wb[0])]=0,i=1;i<tbc;i++) rn[F(wb[i])]=c0(r,wb[i-1],wb[i])?p-1:p++; if(p<tbc) dc3(rn,san,tbc,p); else for(i=0;i<tbc;i++) san[rn[i]]=i; for(i=0;i<tbc;i++) if(san[i]<tb) wb[ta++]=san[i]*3; if(n%3==1) wb[ta++]=n-1; sort(r,wb,wa,ta,m); for(i=0;i<tbc;i++) wv[wb[i]=G(san[i])]=i; for(i=0,j=0,p=0;i<ta && j<tbc;p++) sa[p]=c12(wb[j]%3,r,wa[i],wb[j])?wa[i++]:wb[j++]; for(;i<ta;p++) sa[p]=wa[i++]; for(;j<tbc;p++) sa[p]=wb[j++]; return; } 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++); } void solve() { int i; sum = 0; for (i=1;i<=len1;i++) sum += len1 - sa[i] - height[i]; } int main() { int t,i; while (scanf("%d",&t)!=EOF) { while (t--) { scanf("%s",str1); len1 = strlen(str1); for (i=0;i<len1;i++) { a[i] = static_cast<int>(str1[i]); } a[len1] = 0; dc3(a,sa,len1+1,250); calheight(a,sa,len1); solve(); printf("%d/n",sum); } } return 0; }