[JSOI2007]字符加密

嘟嘟嘟


一看到子串排序，就想办法往后缀数组上靠。


因为后缀数组的排序和长度无关，所以我们把字符串加倍后直接一个后缀排序即可。
然后观察一下输出，其实就是\(sa[i]\)的上一个字符，所以排完序后如果\(sa[i] \leqslant n\)就输出\(s[sa[i] + n - 1]\)即可。
\(O(n \log ^ 2 n)\)的，不开O2还TLE了一个点……

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(‘ ‘)
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 2e5 + 5;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ‘ ‘;
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - ‘0‘, ch = getchar();
  if(last == ‘-‘) ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar(‘-‘);
  if(x >= 10) write(x / 10);
  putchar(x % 10 + ‘0‘);
}

int n, m, K;
char s[maxn];
int sa[maxn], rnk[maxn], tp[maxn];

bool cmp(int i, int j)
{
  if(rnk[i] ^ rnk[j]) return rnk[i] < rnk[j];
  int x = i + K > m ? -1 : rnk[i + K];
  int y = j + K > m ? -1 : rnk[j + K];
  return x < y;
}

int main()
{
  scanf("%s", s + 1);
  n = strlen(s + 1); m = n << 1;
  for(int i = 1; i <= n; ++i) s[i + n] = s[i];
  for(int i = 1; i <= m; ++i) sa[i] = i, rnk[i] = s[i];
  for(K = 1; K <= m; K <<= 1)
    {
      sort(sa + 1, sa + m + 1, cmp);
      for(int i = 1; i <= m; ++i)
    tp[sa[i]] = tp[sa[i - 1]] + (cmp(sa[i- 1], sa[i]) ? 1 : 0);
      for(int i = 1; i <= m; ++i) rnk[i] = tp[i];
    }
  for(int i = 1; i <= m; ++i) if(sa[i] <= n) putchar(s[sa[i] + n - 1]);
  enter;
  return 0;
}