SPOJ NSUBSTR - Substrings

NSUBSTR - Substrings

no tagsYou are given a string S which consists of 250000 lowercase latin letters at most. We define F(x) as the maximal number of times that some string with length x appears in S. For example for string 'ababa' F(3) will be 2 because there is a string 'aba' that occurs twice. Your task is to output F(i) for every i so that 1<=i<=|S|.

Input

String S consists of at most 250000 lowercase latin letters.

Output

Output |S| lines. On the i-th line output F(i).

Example

Input:
ababa

Output:
3
2
2
1
1

解题：SAM求出现次数最多的且长度为i的串的次数

#include <bits/stdc++.h>
using namespace std;
const int maxn = 351000;
struct SAM {
struct node {
int son[26],f,len;
void init() {
f = -1;
len = 0;
memset(son,-1,sizeof son);
}
} s[maxn<<1];
int tot,last;
void init() {
tot = last = 0;
s[tot++].init();
}
int newnode() {
s[tot].init();
return tot++;
}
void extend(int c){
int np = newnode(),p = last;
s[np].len = s[p].len + 1;
while(p != -1 && s[p].son[c] == -1){
s[p].son[c] = np;
p = s[p].f;
}
if(p == -1) s[np].f = 0;
else{
int q = s[p].son[c];
if(s[p].len + 1 == s[q].len) s[np].f = q;
else{
int nq = newnode();
s[nq] = s[q];
s[nq].len = s[p].len + 1;
s[q].f = s[np].f = nq;
while(p != -1 && s[p].son[c] == q){
s[p].son[c] = nq;
p = s[p].f;
}
}
}
last = np;
}
} sam;
char str[maxn];
int c[maxn<<1],ans[maxn],rk[maxn<<1],w[maxn];
int main() {
while(~scanf("%s",str)) {
sam.init();
int len = strlen(str);
for(int i = 0; i < len; ++i)
sam.extend(str[i] - 'a');
memset(ans,0,sizeof ans);
memset(c,0,sizeof c);
memset(w,0,sizeof w);
for(int i = 1; i < sam.tot; ++i) ++w[sam.s[i].len];
for(int i = 1; i <= len; ++i) w[i] += w[i-1];
for(int i = sam.tot-1; i > 0; --i) rk[w[sam.s[i].len]--] = i;
for(int i = 0, p = 0; i < len; ++i)
++c[p = sam.s[p].son[str[i]-'a']];
for(int i = sam.tot-1; i > 0; --i) {
ans[sam.s[rk[i]].len] = max(ans[sam.s[rk[i]].len],c[rk[i]]);
if(sam.s[rk[i]].f > 0) c[sam.s[rk[i]].f] += c[rk[i]];
}
for(int i = 1; i <= len; ++i)
printf("%d\n",ans[i]);
}
return 0;
}

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