题目连接:
https://vjudge.net/problem/SPOJ-NSUBSTR
Description
You 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).
Sample Input
ababa
Sample Output
3
2
2
1
1
Hint
题意
求一个关于串的函数F,F[i]表示串中长度为i的子串出现的最多次数
题解:
这题就是求后缀自动机每个点的Right大小,因为每个endpos类表示的是出现次数及位置一样的一类子串,设一个endpos类中的字符串最长
的长度为len[i],最短为minlen[i], 所以更新时应更新(F[j] = max(F[j], |Right[i]|) j in [minlen[i], len[i]]),但其实不必每个都更新,因为如果长度为n的字符串出现了m次,那么长度为n-1的字符串一定至少出现了m次,及(F[i] >= F[j] (i < j)),所以对于每个点只需要更新(F[len[i]] = max(F[i],|Right[i]|)),最后从大到小更新一下F就好了。
然后就是求解Right大小,先对所有前缀节点赋1,然后自底向上更新right((Right[i] = sum_{edge[i]} Right[j])),为了保证Right正确需要以拓扑序更新Rgiht,也就是len[i]大的节点先更新,因为len[i] > len[fa[i]],len越大的节点越靠近parent tree底部
代码
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int mx = 1e6+5;
struct SAM_automaton {
int Next[mx][26], len[mx], fa[mx];
int last, tot;
int newnode() {
tot++;
for (int i = 0; i < 26; i++) Next[tot][i] = 0;
return tot;
}
void init() {
tot = 0;
last = newnode();
}
void add(int c) {
int p = last;
int np = last = newnode();
len[np] = len[p] + 1;
while (p && !Next[p][c]) {
Next[p][c] = np;
p = fa[p];
}
if (!p) fa[np] = 1;
else {
int q = Next[p][c];
if (len[q] == len[p] + 1) fa[np] = q;
else {
int nq = newnode();
len[nq] = len[p] + 1;
fa[nq] = fa[q];
for (int i = 0; i < 26; i++) Next[nq][i] = Next[q][i];
fa[q] = fa[np] = nq;
while (p && Next[p][c] == q) {
Next[p][c] = nq;
p = fa[p];
}
}
}
}
}SAM;
char str[mx];
int r[mx], b[mx], id[mx], f[mx];
int main() {
SAM.init();
scanf("%s", str);
int len = strlen(str);
for (int i = 0; i < len; i++) SAM.add(str[i] - 'a');
for (int i = 0, p = 1; i < len; i++) {
int c = str[i] - 'a';
r[SAM.Next[p][c]]++;
p = SAM.Next[p][c];
}
for (int i = 1; i <= SAM.tot; i++) b[SAM.len[i]]++;
for (int i = 1; i <= len; i++) b[i] += b[i-1];
for (int i = 1; i <= SAM.tot; i++) id[b[SAM.len[i]]--] = i;
for (int i = SAM.tot; i >= 1; i--) r[SAM.fa[id[i]]] += r[id[i]];
for (int i = 1; i <= SAM.tot; i++) f[SAM.len[i]] = max(f[SAM.len[i]], r[i]);
for (int i = len-1; i >= 1; i--) f[i] = max(f[i], f[i+1]);
for (int i = 1; i <= len; i++) printf("%d
", f[i]);
return 0;
}