POJ 2406 字尾陣列 或 KMP
阿新 • • 發佈:2018-12-24
簡略題意:求解一個連續重複串的最大重複次數。
連續重複串:如果一個字串L由字串S重複R次得到,那麼L是一個連續重複串,R是這個字串的重複次數。
KMP的解法和字尾陣列異曲同工。
不過為了練習字尾陣列,這裡只說字尾陣列的解法。
吐槽一下:
雖然是KMP的水題,不過用字尾陣列寫起來還是很煩的。
1. 卡時間,倍增的字尾陣列過不去。
2. 卡空間,因為串很長,所以不能用一般
假若列舉S串的長度
複雜度
#include <iostream>
#include <cstring>
#include <map>
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
const int N = 1000010;
int n;
namespace SA {
int sa[N], rank[N], height[N], s[N<<1], t[N<<1], p[N], cnt[N], cur[N];
#define pushS(x) sa[cur[s[x]]--] = x
#define pushL(x) sa[cur[s[x]]++] = x
#define inducedSort(v) fill_n(sa, n, -1); fill_n(cnt, m, 0); \
for (int i = 0; i < n; i++) cnt[s[i]]++; \
for (int i = 1; i < m; i++) cnt[i] += cnt[i-1]; \
for (int i = 0; i < m; i++) cur[i] = cnt[i]-1; \
for (int i = n1-1; ~i; i--) pushS(v[i]); \
for (int i = 1; i < m; i++) cur[i] = cnt[i-1]; \
for (int i = 0; i < n; i++) if (sa[i] > 0 && t[sa[i]-1]) pushL(sa[i]-1); \
for (int i = 0; i < m; i++) cur[i] = cnt[i]-1; \
for (int i = n-1; ~i; i--) if (sa[i] > 0 && !t[sa[i]-1]) pushS(sa[i]-1)
void sais(int n, int m, int *s, int *t, int *p) {
int n1 = t[n-1] = 0, ch = rank[0] = -1, *s1 = s+n;
for (int i = n-2; ~i; i--) t[i] = s[i] == s[i+1] ? t[i+1] : s[i] > s[i+1];
for (int i = 1; i < n; i++) rank[i] = t[i-1] && !t[i] ? (p[n1] = i, n1++) : -1;
inducedSort(p);
for (int i = 0, x, y; i < n; i++) if (~(x = rank[sa[i]])) {
if (ch < 1 || p[x+1] - p[x] != p[y+1] - p[y]) ch++;
else for (int j = p[x], k = p[y]; j <= p[x+1]; j++, k++)
if ((s[j]<<1|t[j]) != (s[k]<<1|t[k])) {ch++; break;}
s1[y = x] = ch;
}
if (ch+1 < n1) sais(n1, ch+1, s1, t+n, p+n1);
else for (int i = 0; i < n1; i++) sa[s1[i]] = i;
for (int i = 0; i < n1; i++) s1[i] = p[sa[i]];
inducedSort(s1);
}
template<typename T>
int mapCharToInt(int n, const T *str) {
int m = *max_element(str, str+n);
fill_n(rank, m+1, 0);
for (int i = 0; i < n; i++) rank[str[i]] = 1;
for (int i = 0; i < m; i++) rank[i+1] += rank[i];
for (int i = 0; i < n; i++) s[i] = rank[str[i]] - 1;
return rank[m];
}
template<typename T>
void suffixArray(int n, const T *str) {
int m = mapCharToInt(++n, str);
sais(n, m, s, t, p);
for (int i = 0; i < n; i++) rank[sa[i]] = i;
for (int i = 0, h = height[0] = 0; i < n-1; i++) {
int j = sa[rank[i]-1];
while (i+h < n && j+h < n && s[i+h] == s[j+h]) h++;
if (height[rank[i]] = h) h--;
}
}
int v[N], vv[N];
void solve() {
int rk = rank[0];
v[rk] = N;
for(int i = rk + 1; i <= n; i++)
v[i] = min(v[i-1], height[i]);
for(int i = rk-1; i >= 0; i--)
v[i] = min(v[i+1], height[i+1]);
for(int i = 1; i <= n; i++)
vv[sa[i]] = v[i];
int ans = 1;
for(int i = 1; i <= n; i++) {
if(vv[i] == n-i && n % i == 0)
ans = max(ans, n/i);
}
printf("%d\n", ans);
}
};
int t;
char str[1000000+10];
int main() {
while(~scanf("%s", str) && str[0] != '.') {
n = strlen(str);
SA::suffixArray(n, str);
SA::solve();
}
return 0;
}
/*
aabaabaa
*/