HYSBZ 1031 字符加密Cipher

题目描述：

输入一个字符串，将它的所有循环同构字符串按字典序大小排序，输出这些串最后一个字符依次连接而成的字符串。

题解：

将原串复制一遍并连接，则原串的一个循环同构串是新串的一个长度与原串相等的子串，因此后缀数组排序即可。

题目链接： vjudge 原网站

代码：

#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>using namespace std;#define MAXN 200010static char S[MAXN];static int N;static int sa[MAXN], wa[MAXN], wb[MAXN], wc[MAXN];inline void get_sa(int M){    int *x = wa, *y = wc;    for (int i = 0; i < M; i++) wb[i] = 0;    for (int i = 0; i < N; i++) wb[x[i] = S[i]]++;    for (int i = 1; i < M; i++) wb[i] += wb[i-1];    for (int i = N -1; ~i; i--) sa[--wb[x[i]]] = i;    for (int j = 1, p = 0; p < N; M = p, j <<= 1)    {        p = 0;        for (int i = N - j; i < N; i++) y[p++] = i;        for (int i = 0; i < N; i++) if (sa[i] >= j) y[p++] = sa[i] - j;        for (int i = 0; i < M; i++) wb[i] = 0;        for (int i = 0; i < N; i++) wb[x[y[i]]]++;        for (int i = 1; i < M; i++) wb[i] += wb[i-1];        for (int i = N -1; ~i; i--) sa[--wb[x[y[i]]]] = y[i];        swap(x, y);        x[sa[0]] = 0; p = 1;        for (int i = 1; i < N; i++)            x[sa[i]] = y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+j] == y[sa[i]+j] ? p - 1 : p++;    }}int main(){    scanf("%s", S);    N = strlen(S);    for (int i = 0; i < N; i++) S[i + N] = S[i];    N = 2 * N + 1;    get_sa(300);    for (int i = 0; i < N; i++)        if (sa[i] < (N / 2)) putchar(S[sa[i] + (N / 2) - 1]);    puts("");    return 0;}

提交记录（AC / Total = 1 / 2）：

Run ID Remote Run ID Time(ms) Memory(kb) Result Submit Time 8529516 1941571 764 4140 WA 2017-03-23 07:29:19 8529521 1941581 1160 4140 AC 2017-03-23 07:35:05