[JSOI 2007]字符加密Cipher

Description

题库链接

给你一个长度为 \(n\) 的字符串，首尾相接依次断开每个断点可以得到 \(n\) 个长度为 \(n\) 的字符串，将其排序按序输出每个字符串的最后一个字母。

\(1\leq n\leq 100000\)

Solution

倍长数组直接后缀排序就好了...

Code

#include <bits/stdc++.h>
using namespace std;
const int N = 200000+5;

char ch[N];
int n, m, x[N<<1], y[N<<1], c[N], sa[N];

void get_sa() {
    for (int i = 1; i <= n; i++) ++c[x[i] = ch[i]];
    for (int i = 2; i <= m; i++) c[i] += c[i-1];
    for (int i = n; i >= 1; i--) sa[c[x[i]]--] = i;
    for (int k = 1; k <= n; k <<= 1) {
        int num = 0;
        for (int i = n-k+1; i <= n; i++) y[++num] = i;
        for (int i = 1; i <= n; i++) if (sa[i] > k) y[++num] = sa[i]-k;
        for (int i = 1; i <= m; i++) c[i] = 0;
        for (int i = 1; i <= n; i++) c[x[i]]++;
        for (int i = 2; i <= m; i++) c[i] += c[i-1];
        for (int i = n; i >= 1; i--) sa[c[x[y[i]]]--] = y[i];
        swap(x, y); x[sa[1]] = num = 1;
        for (int i = 2; i <= n; i++)
            x[sa[i]] = (y[sa[i]] == y[sa[i-1]] && y[sa[i]+k] == y[sa[i-1]+k]) ? num : ++num;
        if ((m = num) == n) break;
    }
}
void work() {
    scanf("%s", ch+1); n = strlen(ch+1);
    for (int i = 1; i <= n; i++) ch[i+n] = ch[i];
    n <<= 1, m = 255; get_sa();
    for (int i = 1; i <= n; i++)
        if (sa[i] <= (n>>1)) putchar(ch[sa[i]+(n>>1)-1]);
}
int main() {work(); return 0; }