【LG1368】工艺

【LG1368】工艺

题面

洛谷

题解

好套路的一道题。。。

我们倍长这个字符串，然后我们要查询的串就为这个倍长过后串的长度\(n\)一个子串，要求字典序最小

然后就可以非常愉快地后缀排序了

后缀的话，直接往每个状态的字典序最小的后继状态跑就行了。

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <map>
using namespace std;
inline int gi() {
    register int data = 0, w = 1;
    register char ch = 0;
    while (!isdigit(ch) && ch != '-') ch = getchar();
    if (ch == '-') w = -1, ch = getchar();
    while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar();
    return w * data;
}
const int MAX_N = 3e5 + 5;
struct Node {
    map<int, int> ch;
    int len, fa;
} t[MAX_N << 2];
int tot = 1, lst = 1;
void extend(int c) {
    ++tot, t[lst].ch[c] = tot;
    t[tot].len = t[lst].len + 1;
    int p = t[lst].fa; lst = tot;
    while (p && t[p].ch.find(c) == t[p].ch.end()) t[p].ch[c] = tot, p = t[p].fa;
    if (!p) t[tot].fa = 1;
    else {
        int q = t[p].ch[c];
        if (t[q].len == t[p].len + 1) t[tot].fa = q;
        else {
            int _q = ++tot; t[_q] = t[q];
            t[tot - 1].fa = t[q].fa = _q, t[_q].len = t[p].len + 1;
            while (p) {
                map<int, int> :: iterator ite;
                ite = t[p].ch.find(c);
                if (ite == t[p].ch.end() || ite->second != q) break;
                t[p].ch[c] = _q;
                p = t[p].fa;
            }
        }
    }
}
int N, a[MAX_N];
int main () {
#ifndef ONLINE_JUDGE
    freopen("cpp.in", "r", stdin);
    freopen("cpp.out", "w", stdout);
#endif
    N = gi();
    for (int i = 1; i <= N; i++) a[i] = gi();
    for (int i = 1; i <= N; i++) extend(a[i]);
    for (int i = 1; i <= N; i++) extend(a[i]);
    int v = 1;
    for (int i = 1; i <= N; i++) {
        printf("%d ", t[v].ch.begin()->first);
        v = t[v].ch.begin()->second;
    }
    putchar('\n');
    return 0;
}