POJ 1509 Glass Beads(BOJ 3492,EOlymp 2912,SPOJ BEADS)
http://poj.org/problem?id=1509 https://www.acmicpc.net/problem/3492 https://basecamp.eolymp.com/en/problems/2912 https://www.spoj.com/problems/BEADS/en/
https://vjudge.net/problem/POJ-1509 https://vjudge.net/problem/Baekjoon-3492 https://vjudge.net/problem/EOlymp-2912 https://vjudge.net/problem/SPOJ-BEADS
問題概要
- 文字列$S$が与えられる
- その巡回シフトの中で辞書順最小なのはどこで切ったときか答えよ
- 複数あるなら切る場所の番号が最小のところ
制約
- $|S|\leq 10^4$
解法メモ
-
※ 文字列$S$のSAを求めると、空文字列$\varepsilon$はカウントしないで$0,\cdots,|S|-1$の順列が返ってくるとする
-
辞書順最小のものを見つけるだけなら$S$を2つ繋げてSAを求めて最初に出てくる$N$未満の要素を求めればよい
-
今回はタイブレークまで求められているため、$S+S$ に適当な番兵を入れた文字列のSAを求め、最初に出てくる$N$未満の要素を求めればよい
解法メモ
asciiコードでzの次は{なので$S+S$にこれを連結した文字列のSAを求めている
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
typedef long long ll;
#define rep(i, n) for (ll i = 0, i##_len = (n); i < i##_len; ++i)
using namespace std;
vector<int> sa_is(vector<int> v, int upper) {
// 1 <= v[i] <= upper
if (v.size() == 0) return vector<int>();
else if (v.size() == 1) return vector<int>(1, 0);
else if (v.size() == 2) {
vector<int> res(2, 0);
if (v[0] < v[1]) res[1] = 1;
else res[0] = 1;
return res;
}
v.push_back(0); // sentinel
const int n = v.size();
vector<int> bl(upper + 1), br(upper + 1); // bucket range
for (int i = 0; i < n; ++i) br[v[i]]++;
for (int i = 1; i <= upper; ++i) br[i] += br[i - 1];
for (int i = 1; i <= upper; ++i) bl[i] = br[i - 1];
vector<int> is_l(n);
vector<int> lms, sa(n, -1), lms_ord(n); // lms_ord[i] := 0 -> not lms, 1~ -> 1-indexed
for (int i = n - 2; i >= 0; --i) is_l[i] = (v[i] == v[i + 1]) ? is_l[i + 1] : (v[i] > v[i + 1]);
for (int i = 1; i < n; ++i) {
if (!is_l[i] && is_l[i - 1]) {
sa[--br[v[i]]] = i;
lms_ord[i] = ~int(lms.size());
lms.push_back(i);
}
}
for (int i = 0; i < upper; ++i) br[i] = bl[i + 1];
br[upper] = n;
for (int i = 0; i < n; ++i)
if (sa[i] > 0 && is_l[sa[i] - 1]) sa[bl[v[sa[i] - 1]]++] = sa[i] - 1;
for (int i = 1; i <= upper; ++i) bl[i] = br[i - 1];
for (int i = 1; i < n; i++)
if (sa[i] > -1 && !is_l[sa[i]]) sa[i] = -1;
for (int i = n - 1; i >= 1; i--)
if (sa[i] > 0 && !is_l[sa[i] - 1]) sa[--br[v[sa[i] - 1]]] = sa[i] - 1;
for (int i = 0; i < upper; ++i) br[i] = bl[i + 1];
br[upper] = n;
vector<int> lms_substr_sorted(lms.size());
int cnt = 0;
for (int i = 0; i < n; ++i)
if (sa[i] > -1 && lms_ord[sa[i]]) lms_substr_sorted[cnt++] = sa[i];
// same lms_substr -> same rank
vector<int> ord(lms.size());
ord[0] = 1;
for (int i = 0; i < int(lms.size()) - 1; ++i) {
int l1 = lms_substr_sorted[i], l2 = lms_substr_sorted[i + 1];
if (l1 > l2) swap(l1, l2);
if (l2 == n - 1) ord[i + 1] = ord[i] + 1;
else {
int p1 = l1, p2 = l2;
bool f = true;
while (p1 <= lms[~lms_ord[l1] + 1] && p2 < n)
if (v[p1] == v[p2]) ++p1, ++p2;
else {
f = false;
break;
}
ord[i + 1] = f ? ord[i] : ord[i] + 1;
}
}
vector<int> va(lms.size()); // make array of appearance order
for (int i = 0; i < int(lms.size()); ++i) va[~lms_ord[lms_substr_sorted[i]]] = ord[i];
vector<int> lms_sorted = sa_is(va, ord.back());
// place lms at correct position
fill(sa.begin(), sa.end(), -1);
for (int i = lms.size() - 1; i >= 0; i--) sa[--br[v[lms[lms_sorted[i]]]]] = lms[lms_sorted[i]];
for (int i = 0; i < upper; ++i) br[i] = bl[i + 1];
br[upper] = n;
for (int i = 0; i < n; ++i)
if (sa[i] > 0 && is_l[sa[i] - 1]) sa[bl[v[sa[i] - 1]]++] = sa[i] - 1;
for (int i = 1; i < n; i++)
if (sa[i] > -1 && !is_l[sa[i]]) sa[i] = -1;
for (int i = n - 1; i >= 1; i--)
if (sa[i] > 0 && !is_l[sa[i] - 1]) sa[--br[v[sa[i] - 1]]] = sa[i] - 1;
sa.erase(sa.begin());
return sa;
}
vector<int> sa_is(string s) {
vector<int> v(s.size());
for (int i = 0; i < int(s.size()); i++) v[i] = s[i];
return sa_is(v, 255);
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int q;
cin >> q;
while (q--) {
string s;
cin >> s;
int n = s.size();
s += s;
s += char('z' + 1);
vector<int> sa = sa_is(s);
ll ans = 0;
rep(i, 2 * n) {
if (sa[i] < n) {
cout << sa[i] + 1 << "\n";
break;
}
}
}
}