SPOJ 694 Distinct Substrings

Distinct Substrings

Time Limit: 1000ms

Memory Limit: 262144KB

This problem will be judged on SPOJ. Original ID: DISUBSTR
64-bit integer IO format: %lld      Java class name: Main

 

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.

解题：求不同子串的个数，n - sa[i] - 1表示当前后缀的前缀个数，然后减去 height[i]，也就是减去与前面的重复的。

 #include <bits/stdc++.h>
 using namespace std;
 const int maxn = ;
 int sa[maxn],rk[maxn],height[maxn];
 int c[maxn],t[maxn],t2[maxn],n;
 char s[maxn];
 void build_sa(int m) {
     int i,j,*x = t,*y = t2;
     for(i = ; i < m; ++i) c[i] = ;
     for(i = ; i < n; ++i) c[x[i] = s[i]]++;
     for(i = ; i < m; ++i) c[i] += c[i-];
     for(i = n-; i >= ; --i) sa[--c[x[i]]] = i;

     for(int k = ; k <= n; k <<= ) {
         int p = ;
         for(i = n-k; i < n; ++i) y[p++] = i;
         for(i = ; i < n; ++i)
             if(sa[i] >= k) y[p++] = sa[i] - k;
         for(i = ; i < m; ++i) c[i] = ;
         for(i = ; i < n; ++i) c[x[y[i]]]++;
         for(i = ; i < m; ++i) c[i] += c[i-];
         for(i = n-; i >= ; --i)
             sa[--c[x[y[i]]]] = y[i];
         swap(x,y);
         x[sa[]] = ;
         p = ;
         for(i = ; i < n; ++i)
             if(y[sa[i]] == y[sa[i-]] && y[sa[i]+k] == y[sa[i-]+k])
                 x[sa[i]] = p-;
             else x[sa[i]] = p++;
         if(p >= n) break;
         m = p;
     }
 }
 void getHeight(){
     int i,j,k = ;
     for(i = ; i < n; ++i) rk[sa[i]] = i;
     for(i = ; i < n; ++i){
         if(k) --k;
         j = sa[rk[i]-];
         while(i+k<n&&j+k<n&&s[i+k]==s[j+k]) ++k;
         height[rk[i]] = k;
     }
 }
 int main() {
     int kase;
     scanf("%d",&kase);
     while(kase--){
         scanf("%s",s);
         n = strlen(s) + ;
         build_sa();
         getHeight();
         int ret = ;
         for(int i = ; i < n; ++i)
             ret += n - sa[i] - height[i] - ;
         printf("%d\n",ret);
     }
     return ;
 }

后缀自动机

 #include <bits/stdc++.h>
 using namespace std;
 const int maxn = ;
 char str[maxn];
 int ret,n;
 struct SAM{
     struct node{
         int son[],f,len;
         void init(){
             f = -;
             len = ;
             memset(son,-,sizeof son);
         }
     }e[maxn];
     int tot,last;
     int newnode(int len = ){
         e[tot].init();
         e[tot].len = len;
         return tot++;
     }
     void init(){
         tot = last = ;
         newnode();
     }
     void extend(int c){
         int p = last,np = newnode(e[p].len + );
         while(p != - && e[p].son[c] == -){
             e[p].son[c] = np;
             p = e[p].f;
         }
         if(p == -) e[np].f = ;
         else{
             int q = e[p].son[c];
             if(e[p].len +  == e[q].len) e[np].f = q;
             else{
                 int nq = newnode();
                 e[nq] = e[q];
                 e[nq].len = e[p].len + ;
                 e[q].f = e[np].f = nq;
                 while(p != - && e[p].son[c] == q){
                     e[p].son[c] = nq;
                     p = e[p].f;
                 }
             }
         }
         last = np;
     }
     void solve(){
         ret = ;
         for(int i = ; i < tot; ++i)
             ret += e[i].len - e[e[i].f].len;
         printf("%d\n",ret);
     }
 }sam;

 int main(){
     int kase;
     scanf("%d",&kase);
     while(kase--){
         scanf("%s",str);
         n = strlen(str);
         sam.init();
         for(int i = ret = ; str[i]; ++i)
             sam.extend(str[i]-'A');
         sam.solve();
     }
     return ;
 }