[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 + ;
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 = ; x = ;
   char c = getchar();
   for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
   for (; isdigit(c); c = getchar()) x = (x << ) + (x << ) + c - '';
   x *= f;
}
inline void get_sa()
{
    memset(cnt ,  , sizeof(cnt));
    for (int i = ; i <=  * n; i++) ++cnt[(int)s[i]];
    for (int i = ; i <= ; i++) cnt[i] += cnt[i - ];
    for (int i =  * n; i >= ; i--) sa[cnt[s[i]]--] = i;
    rk[sa[]] = ;
    for (int i = ; i <=  * n; i++) rk[sa[i]] = rk[sa[i - ]] + (s[sa[i]] != s[sa[i - ]]);
    for (int k = ; rk[sa[ * n]] !=  * n; k <<= )
    {
        memset(cnt ,  , sizeof(cnt));
        for (int i = ; i <=  * n; i++)
            x[i] = rk[i] , y[i] = (i + k <=  * n) ? rk[i + k] : ;
        for (int i = ; i <=  * n; i++) ++cnt[y[i]];
        for (int i = ; i <=  * n; i++) cnt[i] += cnt[i - ];
        for (int i =  * n; i >= ; i--) rk[cnt[y[i]]--] = i;
        memset(cnt ,  , sizeof(cnt));
        for (int i = ; i <=  * n; i++) ++cnt[x[i]];
        for (int i = ; i <=  * n; i++) cnt[i] += cnt[i - ];
        for (int i =  * n; i >= ; i--) sa[cnt[x[rk[i]]]--] = rk[i];
        rk[sa[]] = ;
        for (int i = ; i <=  * n; i++) rk[sa[i]] = rk[sa[i - ]] + (x[sa[i]] != x[sa[i - ]] || y[sa[i]] != y[sa[i - ]]);
    }
}

int main()
{

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

    return ;
}