Repeated Substrings

Time Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu

String analysis often arises in applications from biology and chemistry, such as the study of DNA and protein molecules. One interesting problem is to find how many substrings are repeated (at least twice) in a long string. In this problem, you will write a program to find the total number of repeated substrings in a string of at most 100 000 alphabetic characters. Any unique substring that occurs more than once is counted. As an example, if the string is “aabaab”, there are 5 repeated substrings: “a”, “aa”, “aab”, “ab”, “b”. If the string is “aaaaa”, the repeated substrings are “a”, “aa”, “aaa”, “aaaa”. Note that repeated occurrences of a substring may overlap (e.g. “aaaa” in the second case).

The input consists of at most 10 cases. The first line contains a positive integer, specifying the number of
cases to follow. Each of the following line contains a nonempty string of up to 100 000 alphabetic characters.

For each line of input, output one line containing the number of unique substrings that are repeated. You
may assume that the correct answer fits in a signed 32-bit integer.

3
aabaab
aaaaa
AaAaA

5
4
5

解题：后缀数组lcp的应用，如果lcp[i] > lcp[i-1]那么累加lcp[i] - lcp[i-1]

#include
using namespace std;
const int maxn = ;
int rk[maxn],wb[maxn],wv[maxn],wd[maxn],lcp[maxn];
bool cmp(int *r,int i,int j,int k) {
return r[i] == r[j] && r[i+k] == r[j+k];
}
void da(int *r,int *sa,int n,int m) {
int i,k,p,*x = rk,*y = wb;
for(i = ; i < m; ++i) wd[i] = ; for(i = ; i < n; ++i) wd[x[i] = r[i]]++; for(i = ; i < m; ++i) wd[i] += wd[i-]; for(i = n-; i >= ; --i) sa[--wd[x[i]]] = i;

 for(p = k = ; p < n; k <<= ,m = p) {  
     for(p = ,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 < n; ++i) wv\[i\] = x\[y\[i\]\];

     for(i = ; i < m; ++i) wd\[i\] = ;  
     for(i = ; i < n; ++i) wd\[wv\[i\]\]++;  
     for(i = ; i < m; ++i) wd\[i\] += wd\[i-\];  
     for(i = n-; i >= ; --i) sa\[--wd\[wv\[i\]\]\] = y\[i\];

     swap(x,y);  
     x\[sa\[\]\] = ;  
     for(p = i = ; i < n; ++i)  
         x\[sa\[i\]\] = cmp(y,sa\[i-\],sa\[i\],k)?p-:p++;  
 }  

}
void calcp(int *r,int *sa,int n) {
for(int i = ; i <= n; ++i) rk[sa[i]] = i; int h = ; for(int i = ; i < n; ++i) { if(h > ) h--;
for(int j = sa[rk[i]-]; i+h < n && j+h < n; h++) if(r[i+h] != r[j+h]) break; lcp[rk[i]] = h; } } int r[maxn],sa[maxn]; char str[maxn]; int main() { int hn,x,y,cs,ret; scanf("%d",&cs); while(cs--) { scanf("%s",str); int len = strlen(str); for(int i = ; str[i]; ++i) r[i] = str[i]; ret = r[len] = ; da(r,sa,len+,); calcp(r,sa,len); for(int i = ; i <= len; ++i) if(lcp[i] > lcp[i-]) ret += lcp[i] - lcp[i-];
printf("%d\n",ret);
}
return ;
}

后缀自动机

#include
using namespace std;
const int maxn = ;
int cnt[maxn],c[maxn],sa[maxn];
struct node{
int son[],f,len;
void init(){
memset(son,-,sizeof son);
f = -;
len = ;
}
};
struct SAM{
node e[maxn];
int tot,last;
int newnode(int len = ){
e[tot].init();
e[tot].len = len;
return tot++;
}
void init(){
tot = last = ;
newnode();
}
void add(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;
cnt[np] = ;
}
}sam;
char str[maxn];
int main(){
int kase;
scanf("%d",&kase);
while(kase--){
scanf("%s",str);
sam.init();
memset(cnt,,sizeof cnt);
int len = strlen(str);
for(int i = ; str[i]; ++i)
sam.add(str[i]);
node *e = sam.e;
memset(c,,sizeof c);
for(int i = ; i < sam.tot; ++i) c[e[i].len]++; for(int i = ; i <= len; ++i) c[i] += c[i-]; for(int i = sam.tot-; i >= ; --i) sa[--c[e[i].len]] = i;
for(int i = sam.tot-; i > ; --i){
int v = sa[i];
cnt[e[v].f] += cnt[v];
}
int ret = ;
for(int i = ; i < sam.tot; ++i){
if(cnt[i] <= ) continue;
ret += e[i].len - e[e[i].f].len;
}
printf("%d\n",ret);
}
return ;
}