| Time Limit: 1000MS | Memory Limit: Unknown | 64bit IO Format: %lld & %llu |
Description
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.
Source
后缀数组裸题。。。代码真不好记。。所以背下来了。。
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 6 #define maxn 50001 7 #define rep(i,n) for(int i=0;i<n;i++) 8 #define Rep(i,n) for(int i=1;i<=n;i++) 9 #define down(i,n) for(int i=n-1;i>=0;i--) 10 11 int wa[maxn],wb[maxn],cnt[maxn],tx[maxn]; 12 int cmp(int *r,int a,int b,int l){ 13 return r[a]==r[b]&&r[a+l]==r[b+l]; 14 } 15 void BuildSA(int *r,int *sa,int n,int m){ 16 int i,d,p,*x=wa,*id=wb,*t; 17 rep(i,m) cnt[i]=0; 18 rep(i,n) cnt[x[i]=r[i]]++; 19 rep(i,m) cnt[i+1]+=cnt[i]; 20 down(i,n) sa[--cnt[x[i]]]=i; 21 for(d=1,p=1;p<n;d*=2,m=p){ 22 for(p=0,i=n-d;i<n;i++) id[p++]=i; 23 rep(i,n) if(sa[i]>=d) id[p++]=sa[i]-d; 24 rep(i,n) tx[i]=x[id[i]]; 25 rep(i,m) cnt[i]=0; 26 rep(i,n) cnt[tx[i]]++; 27 rep(i,m) cnt[i+1]+=cnt[i]; 28 down(i,n) sa[--cnt[tx[i]]]=id[i]; 29 swap(x,id);p=1;x[sa[0]]=0; 30 Rep(i,n-1) 31 x[sa[i]]=cmp(id,sa[i-1],sa[i],d)?(p-1):(p++); 32 } 33 } 34 int rank[maxn],height[maxn]; 35 void CalHeight(int *r,int *sa,int n){ 36 int i,j,k=0; 37 Rep(i,n) rank[sa[i]]=i; 38 for(i=0;i<n;height[rank[i++]]=k) 39 for(k?k--:0,j=sa[rank[i]-1];r[i+k]==r[j+k];k++); 40 return; 41 } 42 43 char s[maxn]; 44 int r[maxn],sa[maxn]; 45 int main(){ 46 int i,n,t,ans; 47 scanf("%d",&t); 48 while(t--){ 49 scanf("%s",s); 50 n=strlen(s); 51 rep(i,n) r[i]=s[i]; 52 r[n]=0; 53 BuildSA(r,sa,n+1,128); 54 CalHeight(r,sa,n); 55 ans=n*(n+1)/2; 56 Rep(i,n) ans-=height[i]; 57 printf("%d ",ans); 58 } 59 return 0; 60 }