[JSOI 2007] 字符加密

[题目链接]

        https://www.lydsy.com/JudgeOnline/problem.php?id=1031

[算法]

        将字符串倍长 ， 构建后缀数组 ， 然后按要求输出即可

        时间复杂度 : O(NlogN)

[代码]

        

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 10;
typedef long long ll;
typedef long double ld;

int n;
int cnt[MAXN] , sa[MAXN] , rk[MAXN] , x[MAXN] , y[MAXN];
char s[MAXN];

template <typename T> inline void chkmin(T &x , T y) { x = min(x , y); }
template <typename T> inline void chkmax(T &x , T y) { x = max(x , y); }
template <typename T> inline void read(T &x)
{
   T f = 1; x = 0;
   char c = getchar();
   for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
   for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
   x *= f;
}
inline void get_sa()
{
    memset(cnt , 0 , sizeof(cnt));
    for (int i = 1; i <= 2 * n; i++) ++cnt[(int)s[i]];    
    for (int i = 1; i <= 256; i++) cnt[i] += cnt[i - 1];
    for (int i = 2 * n; i >= 1; i--) sa[cnt[s[i]]--] = i;
    rk[sa[1]] = 1;
    for (int i = 2; i <= 2 * n; i++) rk[sa[i]] = rk[sa[i - 1]] + (s[sa[i]] != s[sa[i - 1]]);
    for (int k = 1; rk[sa[2 * n]] != 2 * n; k <<= 1)
    {
        memset(cnt , 0 , sizeof(cnt));
        for (int i = 1; i <= 2 * n; i++)
            x[i] = rk[i] , y[i] = (i + k <= 2 * n) ? rk[i + k] : 0;
        for (int i = 1; i <= 2 * n; i++) ++cnt[y[i]];
        for (int i = 1; i <= 2 * n; i++) cnt[i] += cnt[i - 1];
        for (int i = 2 * n; i >= 1; i--) rk[cnt[y[i]]--] = i;
        memset(cnt , 0 , sizeof(cnt));
        for (int i = 1; i <= 2 * n; i++) ++cnt[x[i]];
        for (int i = 1; i <= 2 * n; i++) cnt[i] += cnt[i - 1];
        for (int i = 2 * n; i >= 1; i--) sa[cnt[x[rk[i]]]--] = rk[i];
        rk[sa[1]] = 1;
        for (int i = 2; i <= 2 * n; i++) rk[sa[i]] = rk[sa[i - 1]] + (x[sa[i]] != x[sa[i - 1]] || y[sa[i]] != y[sa[i - 1]]);
    }
}

int main()
{
    
    scanf("%s" , s + 1);
    n = strlen(s + 1);
    for (int i = 1; i <= n; i++) s[n + i] = s[i];
    get_sa();
    for (int i = 1; i <= 2 * n; i++)
    {
        if (sa[i] + n - 1 < 2 * n) 
            putchar(s[sa[i] + n - 1]);    
    }
    
    return 0;
}